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How to find k'th integer not divisible by n? Although this was a programming question I want the mathematical intuition behind it. So we were given two numbers $n$ and $k$. We were asked to find out $k{-th}$ number **not divisible** by $n$. For example, $n=3$ and
$k=7$.
If we see numbers $1$ to $12$. $$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12$$ Then $3,6,9$ are divisible by $3$ so we remove it. So we are left with $1, 2, 4, 5, 7, 8, 10, 11$. Clearly the $7th$ number not divisible by $3$ is $10$. So answer is $10$. It turns out the answer is:
$$k+⌊\tfrac{k-1}{n-1}⌋$$
I want the mathematical intuition behind this formula.
|
Firstly, we can think about how multiples of $n$ appear in a sequence. For example, when $n=4$:
$$1,2,3,[4],5,6,7,[8],9,10, ...$$
Secondly, we are asked to determine the $k^{th}$ position in the sequence above, "skipping" multiples of $n$. For example when $k=5$, using the same sequence above:
$$1,2,3,[4],5, \langle 6 \rangle ,7,[8],9,10, ...$$
The number $ \langle 6 \rangle $ comes in this natural $6th$ position, but because there's $1$ multiple of $4$ before it, its $k^{th}$ position is deducted by $1$, thus $k^{th}$ of the position $5$ equals $6$.
We can now generalise this counting of multiples by organising them into blocks. This is the spatial reasoning involved in the intuition for the solution:
$$\underline{[1,2,3]},4,\underline{[5,6,7]},8,\underline{[9,10,} ...$$
From the above, clearly, we're counting how many blocks of $3$ (that is $n-1$) precede the numbers that are multiples of $n$, until the $k^{th}$ provided:
$$\lfloor{\frac{k}{n-1}}\rfloor$$
The above will give us how many blocks of numbers, therefore how many multiples, we can "offset" our desired ${k^{th}}$ number, so:
$$k+\lfloor{\frac{k}{n-1}}\rfloor$$
This sort of works, let's see:
$$ \begin{aligned}
k=5, n=4\\
5+\lfloor{\frac{5}{4-1}}\rfloor\\
= 5+\lfloor{\frac{5}{3}}\rfloor\\
= \langle 6 \rangle\\
\end{aligned}
$$
The problem seems to be if $k$ is a multiple of $n-1$ because we will be counting an extra multiple, for example $k=6$, with $n-1 = 3$, we'll have:
$$
k=6, n=4\\
6+\lfloor{\frac{6}{4-1}}\rfloor\\
= 6+\lfloor{\frac{6}{3}}\rfloor\\
= 8 \;\; \text{ WRONG! Correct will be $7$}
$$
Ok, when we hit a block size that divides $k$, we'll have to deduct $1$, since we don't want to count that extra block, so we have:
$$
answer\bigl(k,n \bigr) =
\begin{cases}
k+\lfloor{\cfrac{k}{n-1}}\rfloor - 1 & \text{, if $n-1 \mid k$}\\
k+\lfloor{\cfrac{k}{n-1}}\rfloor & \text{, if $n-1 \nmid k$ }
\end{cases}
$$
This could be the final answer to the problem. It's also $O(1)$ and easy to deconstruct. But we could go further, due to the identities:
$$
\begin{matrix}
\lfloor{\cfrac{m+1}{n}}\rfloor = \lfloor{\cfrac{m}{n}}\rfloor + 1 & \text{, if $n \mid m+1$} & (\alpha)\\
\lfloor{\cfrac{m+1}{n}}\rfloor = \lfloor{\cfrac{m}{n}}\rfloor & \text{, if $n \nmid m+1$} & (\beta)
\end{matrix}
$$
The identities above are easier to reason about, if we prove our intuition using same numbers from our cases above, like so:
$$
\lfloor{\cfrac{5+1}{4-1}}\rfloor = \lfloor{\cfrac{5}{3}}\rfloor + 1 \text{, because $4-1 \mid 5+1$}\\
\lfloor{\cfrac{6+1}{4-1}}\rfloor = \lfloor{\cfrac{6}{3}}\rfloor \text{, because $4-1 \nmid 6+1$}
$$
Or:
$$
\lfloor{\cfrac{6}{3}}\rfloor = \lfloor{\cfrac{5}{3}}\rfloor + 1 \text{, because 3 | 6}\\
\lfloor{\cfrac{7}{3}}\rfloor = \lfloor{\cfrac{6}{3}}\rfloor \text{, because $3 \nmid 7$}
$$
Or:
$$
\begin{matrix}
\lfloor{\cfrac{k}{n-1}}\rfloor = \lfloor{\cfrac{k-1}{n-1}}\rfloor + 1 & \text{, if $n-1 \mid k$} & (\alpha)\\
\lfloor{\cfrac{k}{n-1}}\rfloor = \lfloor{\cfrac{k-1}{n-1}}\rfloor & \text{, if $n-1 \nmid k$} & (\beta)
\end{matrix}
$$
That's where our $k-1$ comes from. So:
$$
k+\lfloor{\frac{k}{n-1}}\rfloor - 1 \text{, if $n-1 \mid k$}\\
\stackrel{\text{from $(\alpha)$}}{=} k+\lfloor{\frac{k-1}{n-1}}\rfloor + 1 - 1\\
= k+\lfloor{\frac{k-1}{n-1}}\rfloor
$$
And:
$$
k+\lfloor{\frac{k}{n-1}}\rfloor \text{, if $n-1 \nmid k$}\\
\stackrel{\text{from $(\beta)$}}{=} k+\lfloor{\frac{k-1}{n-1}}\rfloor
$$
We can combine the two answers now, since both will apply regardless of whether $n-1$ divides $k$ or not:
$$
k+\biggl\lfloor{\frac{k-1}{n-1}}\biggr\rfloor
$$
Q.E.D.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Find $\sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$
For $\displaystyle\omega =
\exp\left({2\pi \over 15}\,\mathrm{i}\right),\quad$ find
$\displaystyle\ \sum_{k=1}^{14} \frac{1}{\left(\omega^{k}-1\right)^{3}}$.
I tried to write $x^{15}-1$=$(x-1) (x-\omega).....(x-\omega^{14})$
And took log and differentiate thrice but it's very lenghty.
|
$w_k$ are the roots of $$y^{15}-1=0$$
Now let $p_k=\dfrac1{(w_k-1)^3}, w_k-1=\sqrt[3]{\dfrac1{p_k}}$
Writing $p_k$ as $z$ and $w_k$ is a root of $y^{15}=1$
$$1=\left(1+\sqrt[3]{\dfrac1z}\right)^{15}$$
$$z^5=(1+\sqrt[3]z)^{15}$$
$$\iff - \sum_{r=0}^4z^r\binom{15}{3r}=z^{1/3}\sum_{r=0}^4 z^r\binom{15}{3r+1}+z^{2/3}\sum_{r=0}^4 z^r\binom{15}{3r+2} $$
Now to rationalize take cube in both sides,
$$-\left(\sum_{r=0}^4z^r\binom{15}{3r}\right)^3=z\left(\sum_{r=0}^4 z^r\binom{15}{3r+1}\right)^3+z^2\left(\sum_{r=0}^4 z^r\binom{15}{3r+2}\right)^3+3z\left(- \sum_{r=0}^4z^r\binom{15}{3r}\right)$$
$$\left(\binom{15}{3\cdot4+2}\right)^3z^{4\cdot3+2}+z^{4\cdot3+1}\left(\left(\binom{15}{3\cdot4+1}\right)^3-3\binom{15}{12}\right)+\cdots=0$$
$$\implies\sum_{k=1}^{14}p_k=-\dfrac{\left(\binom{15}2\right)^3-3\binom{15}{12}}{\left(\binom{15}1\right)^3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Compute the $\lim_{n\to\infty}\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}$ I know you should prove $f_n (x) = \frac1{1+x+\frac{x^n}n}$ converges uniformly and then the limit follows immediately however I’m struggling to prove the uniform convergence.
|
$\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\lt\int_0^1\frac{\mathrm{d}x}{1+x}\\
&= \ln(2)\\
\end{array}
$
and
$\begin{array}\\
f_n
&=\int_0^1\frac{\mathrm{d}x}{1+x+\frac{x^n}n}\\
&\gt\int_0^1\frac{\mathrm{d}x}{1+x+\frac1{n}}\\
&=\ln(2+\frac1{n})-\ln(1+\frac1{n})\\
&=\ln(2)+\ln(1+\frac1{2n})-\ln(1+\frac1{n})\\
&=\ln(2)+\ln(\frac{2n+1}{2n})-\ln(\frac{n+1}{n})\\
&=\ln(2)+\ln(\frac{2n+1}{2(n+1)})\\
&=\ln(2)+\ln(1-\frac1{2(n+1)})\\
&>\ln(2)-\dfrac{\frac1{2(n+1)}}{1-\frac1{2(n+1)}}
\qquad\text{since } \ln(1-x) > -\dfrac{x}{1-x} (*)\\
&=\ln(2)-\dfrac{1}{2(n+1)-1}\\
&=\ln(2)-\dfrac{1}{2n+1}\\
\end{array}
$
so
$f_n \to \ln(2)$.
$(*)\ -\ln(1-x)
=\int_{1-x}^1 \dfrac{dx}{x}
\lt \dfrac{x}{1-x}
$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Sum of Ranks of Two Complementary Matrices Suppose we have a binary matrix $A$, i.e, all the elements of $A$ are either $0$ or $1$. Let $B$ be the complementary matrix of $A$, i.e.,
$$
B_{ij}=
\begin{cases}
1,\textrm{ if }A_{ij}=0;\\
0,\textrm{ if }A_{ij}=1.
\end{cases}
$$
Then $A+B$ is a matrix with all entries being $1$. Suppose $\textrm{rank}(A)=r$, what is $\textrm{rank}(A)+\textrm{rank}(B)$?
|
There is not enough information to tell.
For example, let $s=\operatorname{rank} B$. If $n=2$, we can take
*
*$A=\begin{bmatrix} 1&1\\0&0\end{bmatrix}$, then $r+s=2$
*$A=\begin{bmatrix} 1&0\\0&0\end{bmatrix}$, then $r+s=3$
*$A=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$, then $r+s=4$
Similarly, with $n=3$:
*
*$A=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0\end{bmatrix}$, then $r+s=3$
*$A=\begin{bmatrix} 1&0&1\\0&1&0\\0&0&0\end{bmatrix}$, then $r+s=4$
*$A=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&0\end{bmatrix}$, then $r+s=5$
*$A=\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1\end{bmatrix}$, then $r+s=6$
|
{
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"url": "https://math.stackexchange.com/questions/3674871",
"timestamp": "2023-03-29T00:00:00",
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|
Primes of the form $x^2 + 9 y^2$ and $x^2 + 12 y^2$ I've been studying primes of the form $x^2 + n y^2$ (where of course $x$, $y \in \mathbb{Z}$) and I noticed the following:
$$p = x^2 + 9 y^2 \iff p \equiv 1 \,\, (\textrm{mod} \,\, 12) \iff p = u^2 + 12 v^2$$
For example, $10^2 + 9 \cdot 7^2 = 541 = 23^2 + 12 \cdot 1^2$.
The "difficult halves" of these biconditionals can be gotten immediately from the fact that
an integer $m$ is represented by a quadratic form of discriminant $\Delta$ $\iff$ $\Delta$ is a square mod $4m$
The reduced quadratic forms of discriminant $-36$ are $x^2 + 9 y^2$ and $3 x^2 + 3 y^2$ and $2 x^2 + 2 x y + 5 y^2$. If $p \equiv 1 \,\, (\textrm{mod} \,\, 12)$, then $-36$ is a square mod $4p$, and among these three forms, the only one with any $1 \,\, (\textrm{mod} \,\, 3)$ outputs is $x^2 + 9 y^2$.
Likewise, the reduced quadratic forms of discriminant $-48$ are $u^2 + 12 v^2$ and $2 u^2 + 6 v^2$ and $3 u^2 + 4 v^2$. If $p \equiv 1 \,\, (\textrm{mod} \,\, 12)$, then $-48$ is a square mod $4p$; and among these three forms, the only one with any $1 \,\, (\textrm{mod} \,\, 4)$ outputs is $u^2 + 12 v^2$.
So $p \equiv 1 \,\, (\textrm{mod} \,\, 12)$ implies both $p = x^2 + 9 y^2$ and $p = u^2 + 12 v^2$.
My question: is it just a coincidence that $x^2 + 9 y^2$ and $u^2 + 12 v^2$ represent exactly the same primes?
I've computed $(x,\,y)$ and $(u,\,v)$ for the relevant primes under $1000$, and there doesn't seem to be a pattern relating them.
Thanks!
|
Not a concidence.
I will prove that if a prime p is of the form $x^2+9y^2$ then it must be of the form $a^2+12b^2$ and viceversa.
Case 1: $p=a^2+12b^2$ Prove $p=x^2+ 9y^2$ for some $x$ and $y$
$p=a^2+12b^2$ Note that $p$ is $\equiv 1\pmod{4}$ so -1 is also a quadratic residue, so $-1\times9=9$ also is one.
From Thue's lemma we have, by taking $a$ such that $a^2\equiv-9\pmod{p}$ (obviously, $p\neq 3$) some $x$, $y$ such that $x\equiv ay\pmod{p}$ so $x^2\equiv -9y^2\pmod{p}$ so $x^2+9y^2$ is divisible by p. Also from Thue's lemma we have that $x^2+9y^2<10p$ Now a simole analysis will lead to the fact that $p$ is indded of the form $x^2+9y^2$
Case 2: $p=x^2+ 9y^2$ Prove $p=a^2+12b^2$ for some $a$ and $b$
$x\equiv 1\pmod{3}$ so $\big(\frac{-3}{p}\big)=\big(\frac{3}{p}\big)\times\big(\frac{-1}{p}\big)=\frac{1}{\big(\frac{p}{3}\big)}\times (-1)^{\frac{p-1}{2}\frac{3-1}{2}}\times(-1)^{\frac{p-1}{2}}=1$ From the quadratic reciprocity law so $-3$ is a qudratic residue. Again use Thue to prove that there exist $a$ and $b$ such that $a^2+12b^2$ is divisible by $p$ and $<13p$ and from a simple analysis we get that $p$ is indeed of the form we wanted
So we win!
|
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"timestamp": "2023-03-29T00:00:00",
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|
Absolute minimum and maximum of $f(x,y,z)=x^4+y^4+z^4-4xyz$ I have to find the absolute maximum and minimum of the function $f(x,y,z)=x^4+y^4+z^4-4xyz$ over $x^2+y^2+z^2\leq 9$, $x,y,z\geq 0$.
I'm having problems to find the constrained extremas in $x^2+y^2+z^2=9$. I have tried by using the Lagrange's Multipliers theorem, by parametrizing the sphere and by algebraic manipulating the function but I haven't been able to come up with the solution.
Could anybody help me?
Thank you in advance.
|
Integrand's concern about the case for $ \ xyz \ \ne \ 0 \ $ turns out to be unnecessary. If we solve the original Lagrange equations for $ \ \lambda \ $ , we obtain
$$ \left\{ \begin{array}{c}
2·(x^3-yz) \ = \ \lambda x \\ 2·(y^3-4xz) \ = \ \lambda y \\
2·(z^3- xy) \ = \ \lambda z
\end{array} \right. \ \ \Rightarrow \ \ \lambda \ = \ \frac{2·(x^3-yz)}{x} \ = \ \frac{2·(y^3-xz)}{y} \ = \ \frac{2·(z^3-xy)}{z} \ \ . $$
Cross-multiplying the first ratio-pair produces
$$ x^3y \ - \ y^2z \ \ = \ \ xy^3 - x^2z \ \ \Rightarrow \ \ x^2z \ - \ y^2z \ \ = \ \ xy^3 - x^3y $$ $$ \Rightarrow \ \ z·(x^2-y^2) \ \ = \ \ xy·(y^2 - x^2) \ \ \Rightarrow \ \ (z \ + \ xy) \ · \ (x^2-y^2) \ \ = \ \ 0 \ \ ; $$
similarly, we have from the other ratio-pairs,
$$ (y \ + \ xz) \ · \ (x^2-z^2) \ \ = \ \ 0 \ \ \ , \ \ \ (x \ + \ yz) \ · \ (y^2-z^2) \ \ = \ \ 0 \ \ . $$
[That each of these equations looks like a "cyclic-permutation" re-labeling of the others is due to the same symmetry being found in the expression for $ \ f(x,y,z) \ $ and the symmetry of the spherical constraint.]
The "difference-of-squares" factors would tell us such things as $ \ y = \pm x \ , $ but since we are also restricted to the first octant, we have just $ \ x = y = z \ $ as one case. This gives the $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ results.
The remaining case comes from the factors such as $ \ z + xy \ = \ 0 \ \Rightarrow \ z \ = \ -xy \ . $ Again because we are in the first octant, including those portions of the coordinate planes, this requires that $ \ z = 0 \ $ and either $ \ x = 0 \ \ \text{or} \ \ y=0 \ . $ Together with the related results from the other two equations, we obtain the three critical points which are the permutations of $ \ (3 \ , \ 0 \ , \ 0 ) \ $ and the three permutations of $ \ (3\sqrt{2} \ , \ 3\sqrt{2} \ , \ 0 ) \ . $ So it does not appear that any critical points have been overlooked and the conclusion in Integrand's 19 May post is confirmed.
(Incidentally, the origin is a rather interesting saddle point. The function $ \ f(x,x,x) \ = \ 3x^4 - 4x^3 \ $ has an inflection point at $ \ x = 0 \ $ , so this change in concavity occurs along each of the lines passing between opposing octants:
$$ -x = -y = -z \ \rightarrow \ x = y = z \ \ , \ \ x = -y = -z \ \rightarrow \ -x = y = z \ \ , $$
$$ -x = y = -z \ \rightarrow \ x = -y = z \ \ \text{and} \ \ -x = -y = z \ \rightarrow \ x = y = -z \ \ . ) $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that there's no ordered basis $E$ in which $T{x\choose y}={0\choose y}$ can be represented as $1\ 2\choose 2\ 4$ Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^2$ such that:
$$T\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix}0\\ y\end{pmatrix}$$
Show that there's no ordered-basis, $E$ such that:
$$[T]_E = \begin{pmatrix}
1 & 2\\
2 & 4\\
\end{pmatrix}$$
I didn't get how should I prove it. I tried some ways but end up with nothing. I guess I'm missing something.
|
With
$T \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ y \end{pmatrix}, \tag 1$
we have
$T \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \tag 2$
and
$T \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \tag 3$
which shows the eigenvalues of $T$ are $0$ and $1$; on the other hand, the eigevalues of the matrix
$[T]_E = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \tag 4$
are the roots of its characteristic polynomial
$\det ([T]_E - \lambda I) = \begin{pmatrix} 1 - \lambda & 2 \\ 2 & 4 - \lambda \end{pmatrix}$
$= (1 - \lambda)(4 - \lambda) - 4 = 4 - 5\lambda + \lambda^2 - 4 = \lambda^2 - 5 \lambda, \tag 5$
which are $0$ and $5$; since the eigenvalues of the transformation $T$ as in (1) are $0$ and $1$, it cannot be represented by (4) in any basis $E$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
When converting a Riemann sum to an integral, how does one decide the bounds of the converted Riemann sum? Context problem: $$ \lim_{n\to \infty} \frac{1}{n} \sum_{r=1}^{r=2n} \frac{r}{\sqrt{r^2+n^2}}$$
So the thing messing me up is the '2n' in the upper of sum, I've already figured out that the function which this rienman integral is ( $ \int \frac{x}{\sqrt{1+x^2}}$)
|
And now a synthesis of the other two answers that involves less (apparent) guessing than one and slightly more precision than the other.
Suppose what you are given is a Riemann sum similar to the ones you performed by hand when introduced to the concept of Riemann sums (left Riemann sum, right Riemann sum, midpoint ...) Then you expect that you are looking at an expression of the form
$$ \lim_{\text{the partition shrinks}} \Delta x \sum \text{sample heights} $$
So provisionally, we take $\Delta x = \frac{1}{n}$. (Provisionally, because our subsequent algebraic manipulations may result in modifying this term.)
The sum you are given uses $r$ as the index, so $r$ is which member of the partition we are sampling. We have $\Delta x = \frac{1}{n}$, so we expect the sample points to be (not necessarily exactly) $1/n$ apart, so sampled at $r$ steps of size $1/n$, which is $r/n$. So we want to algebraically manipulate the summand so that all occurrences of $r$ are of the form $r/n$. \begin{align*}
\frac{r}{\sqrt{r^2 + n^2}}
&= \frac{n \cdot r/n}{\sqrt{n^2 \cdot (r/n)^2 + n^2}} \\
&= \frac{n \cdot r/n}{\sqrt{n^2 \cdot \left( (r/n)^2 + n^2/n^2 \right)}} \\
&= \frac{n \cdot r/n}{n \sqrt{(r/n)^2 + 1 }} \\
&= \frac{r/n}{\sqrt{(r/n)^2 + 1 }} \text{.}
\end{align*}
(Technicality: in the denominator, we wrote $\sqrt{n^2} = n$. This isn't quite right. $\sqrt{n^2} = |n|$. However, we have $0 < n$, so $|n| = n$ here.)
Having succeeded, we know that the sample points of the function $\frac{x}{\sqrt{x^2+1}}$ occur at $x = r/n$. (There is a comment about horizontal translation and scaling below that may be of use to some readers at this point.) And these are separated by $1/n$, so $\Delta x = 1/n$ is promoted from provisionally correct to correct. If this were not the case, our job would be to find $p$ so that $\Delta x = 1/n^p$ appears in front of the sum and $r$ appears in the combination $r/n^p$ throughout. (This depends on the fact that the summand/integrand is an algebraic fraction, so such a $p$ can be found. If the integrand is more complicated, $\Delta x$ and the combination in which $r$ appears may be more complicated.)
But now we know the endpoints. Since $x$ is $r/n$, and $r$ ranges from $1$ to $2n$, we need to take the limits of the left-most $x$, $1/n$ and the right-most $x$, $2n/n$:
\begin{align*}
\lim_{n \rightarrow \infty} 1/n &= 0 \text{ and } \\
\lim_{n \rightarrow \infty} 2n/n &= 2 \text{.}
\end{align*}
From which, we know that an integral equivalent to our sum is
$$ \int_0^2 \frac{x}{\sqrt{x^2 + 1}} \,\mathrm{d}x \text{.} $$
Why do I say "an"? Because there are many integrals equivalent to our sum. Here's another.
$$ \int_7^9 \frac{u-7}{\sqrt{u^2 - 14u + 50}} \,\mathrm{d}u \text{,} $$
which is obtained by translating the previous integral to the right by $7$ units. That is, by the substitution $u = x+7$. We can make this integral be the one we land on by making $\frac{r}{n} + 7$ be the combination in which $r$ appears.
\begin{align*}
\frac{r}{\sqrt{r^2 + n^2}}
&= \frac{n \cdot ((r/n+7) - 7)}{\sqrt{n^2 \cdot ((r/n+7) - 7)^2 + n^2}} \\
&= \frac{n \cdot ((r/n+7) - 7)}{\sqrt{n^2 \cdot \left( ((r/n+7) - 7)^2 + n^2/n^2 \right)}} \\
&= \frac{n \cdot ((r/n+7) - 7)}{n \cdot \sqrt{ ((r/n+7) - 7)^2 + 1}} \\
&= \frac{(r/n+7) - 7}{\sqrt{(r/n+7)^2 - 14(r/n+7) + 50}} \text{.}
\end{align*}
With $x = \frac{r}{n} + 7$, the integrand and limits of integration are
$$ \frac{x-7}{\sqrt{x^2 - 14x + 50}} \text{,} $$
\begin{align*}
\lim_{n \rightarrow \infty} \frac{1}{n} + 7 &= 7 \text{, and } \\
\lim_{n \rightarrow \infty} \frac{2n}{n} + 7 &= 9 \text{.}
\end{align*}
We can even scale $x$ to $kr/n$ for $k \neq 0$. Similarly to the translation, we algebraically force the combination of $r$ to match. Unlike translation, our $\Delta x$ changes to $k/n$. (This should be unsurprising. When you make the substitution $u = kx$, you scale the variable in the integrand and scale the differential element.)
I mention translation and scaling to remove a common worry by students. There are many integrals corresponding to the sum you are given differing by scaling and translation. You only have to find one such integral. The integral that is easiest to find has $r$ appearing in a simple combination (no translation) and $\Delta x$ appearing without a constant multiplier (no scaling). You can always find the "simpler" integral (lacking translation or scaling) using the procedure we described first.
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $N = y^2+1$ then the continued fraction of $\sqrt(N)$ has period 1. I am reading a book that says that if $N = y^2+1$ then the continued fraction of $\sqrt{N}$ has period 1, i.e $\sqrt{N} = [q_0;\overline{q_1}]$ or similarly
$$x=q_0+\cfrac{1}{q_1+\cfrac{1}{q_1+\cfrac{1}{ q_1+ \cdots}}}$$
What I've tried to do was to compute the continued fraction of $\sqrt{y^2+1}$ directly:
Therefore $q_0 = \lfloor \sqrt{y^2+1} \rfloor = y$ and I tried to compute $$\frac{1}{\sqrt{y^2+1}-y} = \frac{\sqrt{y^2+1}-y}{2y^2+1-2y\sqrt{y^2+1}} = \frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)+1} < \frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)}$$
But here I am stuck and I guess that I made an stupid error because there's no way that $$\frac{\sqrt{y^2+1}-y}{-2y(\sqrt{y^2+1}-y)+1} < \frac{1}{-2y}$$
|
$\begin{array}\\
\sqrt{y^2+1}-y
&=(\sqrt{y^2+1}-y)\dfrac{\sqrt{y^2+1}+y}{\sqrt{y^2+1}+y}\\
&=\dfrac{1}{\sqrt{y^2+1}+y}\\
\end{array}
$
so
$\sqrt{y^2+1}
=y+\dfrac{1}{y+\sqrt{y^2+1}}
$.
|
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|
Prove that for all $\alpha + \beta + \gamma = \pi$, $\sum_{cyc}\frac{\sin\beta}{\cos\beta + 1} = \frac{\sum_{cyc}\cos\beta + 3}{\sum_{cyc}\sin\beta}$.
Prove that for all triangles with angles $\alpha, \beta, \gamma$, $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 3}{\sin\alpha + \sin\beta + \sin\gamma}$$
Let $\tan\dfrac{\alpha}{2} = a, \tan\dfrac{\beta}{2} = b, \tan\dfrac{\gamma}{2} = c$, we have that $$\dfrac{\sin\beta}{\cos\beta + 1} = \dfrac{1}{b}, \cos\beta = \dfrac{1 - b^2}{1 + b^2}, \sin\beta = \dfrac{2b}{1 + b^2}$$ and $bc + ca + ab = 1$.
It needs to be proven that $$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1 - a^2}{1 + a^2} + \dfrac{1 - b^2}{1 + b^2} + \dfrac{1 - c^2}{1 + c^2} + 3}{\dfrac{2a}{1 + a^2} + \dfrac{2b}{1 + b^2} + \dfrac{2c}{1 + c^2}}$$
$$\impliedby \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{\dfrac{1}{1 + a^2} + \dfrac{1}{1 + b^2} + \dfrac{1}{1 + c^2}}{\dfrac{a}{1 + a^2} + \dfrac{b}{1 + b^2} + \dfrac{c}{1 + c^2}}$$
$$\impliedby \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)\left(\frac{a}{1 + a^2} + \frac{b}{1 + b^2} + \frac{c}{1 + c^2}\right) = \frac{1}{1 + a^2} + \frac{1}{1 + b^2} + \frac{1}{1 + c^2}$$
$$\impliedby \left(\frac{a}{b} + \frac{a}{c}\right)\frac{1}{1 + a^2} + \left(\frac{b}{c} + \frac{b}{a}\right)\frac{1}{1 + b^2} + \left(\frac{c}{a} + \frac{c}{b}\right)\frac{1}{1 + c^2} = 0$$
$$\impliedby \frac{a(b + c)}{bc(c + a)(a + b)} + \frac{b(c + a)}{ca(a + b)(b + c)} + \frac{c(a + b)}{ab( b + c)(c + a)} = 0$$
$$\impliedby \frac{1 - bc}{(1 - ca)(1 - ab)} + \frac{1 - ca}{(1 - ab)(1 - bc)} + \frac{1 - ab}{(1 - bc)(1 - ca)} = 0$$
$$\impliedby (1 - bc)^2 + (1 - ca)^2 + (1 - ab)^2 = 0$$
$$\impliedby bc = ca = ab = 1 \impliedby bc + ca + ab = 3,$$ which is definitely incorrect.
I've surmised that the correct equality is $$\frac{\sin\alpha}{\cos\alpha + 1} + \frac{\sin\beta}{\cos\beta + 1} + \frac{\sin\gamma}{\cos\gamma + 1} = \frac{\cos\alpha + \cos\beta + \cos\gamma + 1}{\sin\alpha + \sin\beta + \sin\gamma},$$ but then I wouldn't know what to do first.
|
For $1+\cos\alpha\ne0,$
$$\dfrac{\sin\alpha}{1+\cos\alpha}=\cdots=\tan\dfrac\alpha2$$
Now,
$$\tan\dfrac\alpha2+\tan\dfrac\beta2+\tan\dfrac\gamma2$$
$$=\dfrac{\sin\left(\dfrac\alpha2+\dfrac\beta2\right)}{\cos\dfrac\alpha2\cos\dfrac\beta2}+\dfrac{\sin\dfrac\gamma2}{\cos\dfrac\gamma2}$$
$$=\dfrac{\cos\dfrac\gamma2}{\cos\dfrac\alpha2\cos\dfrac\beta2}+\dfrac{\sin\dfrac\gamma2}{\cos\dfrac\gamma2}\text{ using }\alpha+\beta=\pi-\gamma$$
$$=\dfrac{\cos^2\dfrac\gamma2+\sin\dfrac\gamma2\cos\dfrac\alpha2\cos\dfrac\beta2}{\cos\dfrac\alpha2\cos\dfrac\beta2\cos\dfrac\gamma2}$$
Now the numerator
$$= 1-\sin^2\dfrac\gamma2+\sin\dfrac\gamma2\cos\dfrac\alpha2\cos\dfrac\beta2 $$
$$= 1-\sin\dfrac\gamma2\left(\sin\dfrac\gamma2-\cos\dfrac\alpha2\cos\dfrac\beta2\right) $$
$$= 1-\sin\dfrac\gamma2\left(\cos\left(\dfrac\alpha2+\dfrac\beta2\right) -\cos\dfrac\alpha2\cos\dfrac\beta2\right) $$
$$= 1+\sin\dfrac\alpha2\sin\dfrac\beta2\sin\dfrac\gamma2$$
Now use Prove trigonometry identity for $\cos A+\cos B+\cos C$
and If $A + B + C = \pi$, then show that $\sin(A) + \sin(B) + \sin(C) = 4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\gcd\left(n^{a}+1, n^{b}+1\right)\mid (n^{\gcd(a, b)}+1)$
Let $a$ and $b$ be positive integers. Prove that $\operatorname{gcd}\left(n^{a}+1, n^{b}+1\right)$ divides $n^{\operatorname{gcd}(a, b)}+1$.
My work -
I proved this for $n=2$ but I am not able to prove this for all $n$ (if anyone wants I can give my proof for $n=2$).
More Observation.
If $a$ and $b$ are both odd, then $d=\gcd(a,b)$ is an odd positive integer. Therefore,
$$n^a+1=(n^d+1)\left(n^{d(a-1)}-n^{d(a-2)}+\ldots-n^d+1\right)$$
and
$$n^b+1=(n^d+1)\left(n^{d(b-1)}-n^{d(b-2)}+\ldots-n^d+1\right),$$
whence $n^d+1$ divides both $n^a+1$ and $n^b+1$. That is, $n^d+1$ divides $\gcd(n^a+1,n^b+1)$. However, we can perform Euclidean algorithm as follows.
Without loss of generality, let $a\geq b$.
Case I: $a\geq 2b$. We have
$$n^a+1=(n^{b}+1)\left(n^{a-b}-n^{a-2b}\right)+(n^{a-2b}+1)\,.$$
We can replace $(a,b)$ by $(a-2b,b)$, and perform more reduction steps.
Case II: $b<a<2b$. We have
$$n^{a}+1=(n^b+1)n^{a-b}-\left(n^{a-b}-1\right)$$
and
$$n^b+1=\left(n^{a-b}-1\right)n^{2b-a}+(n^{2b-a}+1)\,.$$
Thus, we can replace $(a,b)$ by $(b,2b-a)$ and perform more reduction steps.
Case III: $a=b$. Then, the reduction steps end.
Note that, at each step, the difference between $a$ and $b$ never increases. (Observe that, we cannot perform the steps in Case II infinitely many times, as the smaller value between $a$ and $b$ always decreases.) Therefore, the process has to stop when both numbers become the same odd integer $s$, which is an integer combination of $a$ and $b$. However, $d$ divides any integer combination of (the starting values of) $a$ and $b$. Thus, $d$ divides $s$. The Euclidean algorithm above shows that $n^s+1$ is the greatest common divisor of $n^a+1$ and $n^b+1$. Thus, $s=d$, so in the case $a$ and $b$ are odd,
$$\gcd(n^a+1,n^b+1)=n^{\gcd(a,b)}+1\,.$$
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Suppose that for some prime $p$ and positive integer $k$ we have $p^k$ divides both $n^a+1$ and $n^b+1$. Then, we need to prove that $p^k$ divides $n^{\gcd(a,b)}+1$. Denote $d=\gcd(a,b)$. Here, we will consider two cases:
Case 1. $p=2$. In this case, if $a$ or $b$ is even, then $k=1$ (because $m^2+1$ can't be divisible by 4) and $n$ should be odd. So, $n^d-1$ is divisible by $p^k=2$, as desired.
If both $a$ and $b$ is odd, then $\gcd(n^a+1, n^b+1)=n^d+1$ (it's similar to Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$) and in particular, $2^k\mid n^d+1$.
Case 2. $p>2$. In this case, note that $p^k$ divides $$n^{2a}-1=(n^a-1)(n^a+1)$$ and $$n^{2b}-1=(n^b-1)(n^b+1)\,,$$ so $p^k$ divides $n^{2d}-1=(n^d-1)(n^d+1)$. Note that $p$ can't divide both $n^d-1$ and $n^d+1$ (because $p>2$). Hence, it's sufficient to prove that $n^d-1$ can't be divisible by $p^k$. Indeed, if $n^d\equiv 1\pmod {p^k}$, then $$n^a\equiv n^b\equiv 1\pmod {p^k}\,.$$ However, by our assumption we have $n^a\equiv n^b\equiv -1\pmod {p^k}$, so due to $p^k>2$ we get a contradiction. Thus, $p^k$ divides $n^d+1$ as desired.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $g(x)$ if $f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$ and $g(x) = |f(x)| + f(|x|)$ $$f:[-2,2] \rightarrow \Bbb R$$
$$\text {and }f(x)= \begin{cases} -1, & -2 \leq x \leq 0 \\ x-1, & 0 < x \leq 2 \end{cases}$$
And, let $g(x)$ be equal to $|f(x)|+f(|x|)$
We need to find the value of $g(x)$ (define it).
I begin by finding the value of $|f(x)|$ first :
$$|f(x)|=\begin{cases} |-1|, & -2 \leq x \leq 0 \\ |x-1|, & 0 < x \leq 2 \end{cases}=\begin{cases} 1, & -2 \leq x \leq 0 \\ |x-1|, & 0 < x \leq 2\end{cases}$$
Now, to determine what $|x-1|$ would be, we need to determine whether $(x-1)$ is positive or negative or zero.
If $x-1 \geq 0$, $|x-1| = x-1$ and if $x-1 < 0$, then $|x-1| = -(x-1) = 1-x$
If $x-1 \geq 0$, then $x \geq 1$ and if $x - 1 < 0$, then $x < 1$. We can now split the condition $0 < x \leq 2$ from the earlier definition of $|f(x)|$ as $0<x<1$ and $1 \leq x \leq 2$, where $0<x<1 \implies |x-1| = 1-x$ and $1 \leq x \leq 2 \implies |x-1| = x-1$
So, $$ |f(x)| = \begin{cases} 1, & -2 \leq x \leq 0 \\ 1-x, & 0<x<1 \\ x-1, & 1 \leq x \leq 2 \end{cases}$$
Now, we need to define $f(|x|)$. Here's what I do :
$$f(|x|) = \begin{cases} -1, & -2 \leq x \leq 0 \\ |x|-1, & 0<|x| \leq 2 \end{cases}$$
Now, is the next step that I do here correct or even necessary?
Now, $|x|$ can never be negative but can be zero when $x=0$. So, the condition $-2 \leq |x| \leq 0$ can be replaced by $x = 0$ from which we obtain the following definition for $f(|x|)$:
$$f(|x|) = \begin{cases} -1, & x=0 \\ |x|-1, & 0 < x \leq 2 \end{cases} = \begin{cases} |x|-1, & 0 \leq |x| \leq 2 \end{cases}$$
I did the last one because we observe that $0$ is mapped to $-1$ and $|0|-1 = -1$, so it will still be mapped to $-1$ if we put it in the second condition. So, basically, for any value of $x$ that is a part of domain of $f$, $f(|x|) = |x|-1$.
Now, we add $|f(x)|$ and $f(|x|)$ to obtain $g(x)$.
$$g(x) = |f(x)| + f(|x|) = \begin{cases} |x|-1+1, & -2 \leq x \leq 0 \\ |x|-1+1-x, & 0 < x < 1 \\ |x|-1+x-1, & 1 \leq x \leq 2\end{cases}$$.
In the first condition, the value of $x$ is either negative or $0$, so if it is negative, $|x| = -x$ and else, it is 0 which is also equal to $-0$, which means that in the first condition, $|x|$ can be substituted by $-x$. In the second and, the value of $x$ is always positive, so $|x| = x$. So, we can substitute $|x|$ by $x$ in the second and third conditions and arrive at the definition of $g(x)$ that is :
$$g(x) = \begin{cases} -x-1+1, & -2 \leq x \leq 0 \\ x-1+1-x, & 0 < x < 1 \\ x-1+x-1, & 1 \leq x \leq 2 \end{cases} = \begin{cases} -x, & -2 \leq x \leq 0 \\ 0, & 0 < x < 1 \\ 2x-2, & 1 \leq x \leq 2 \end{cases}$$
Now, I want to know if this process is correct and if there is some alternative, better approach to this problem.
Sorry for the long post, I thought that showing my line of reasoning would make the question better.
Thanks
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I used Desmos to draw the graphs:
$$f(x)$$
$$|f(x)|$$
$$f(|x|)$$
$$g(x)$$
While I love desmos, I think you solution for f(|x|) is more realistic.
And then, the graph of g(x) is:
Notice that because the graph f(|x|) is not defined at x<0, then g(x) is not defined there.
So, in conclusion, $$g(x) = \begin{cases}
0&x\le1\\
2(x-1)&1<x\le2
\end{cases}$$
|
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|
Prove that $\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$ with $a>0, b>0 , c> 0$ and $d>0.$
Prove that
$\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)\geq4$
with $a>0, b>0 , c> 0$ and $d>0.$
My attempt:
$$\begin{align*}\left(\dfrac{b}{a}+\dfrac{d}{c}\right)\cdot\left(\dfrac{a}{b}+\dfrac{c}{d}\right)& = \dfrac{abcd+b^2c^2+a^2d^2+abcd}{abcd}\\
& =\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{b^2c^2+a^2d^2+2abcd}{abcd}\\
&=\dfrac{(ad)^2+(bc)^2+2(ad)(bc)}{abcd}\\
&=\dfrac{(ad+bc)^2}{abcd}\end{align*}$$
I don't know how to continue from this.
Can someone help me?
|
Let $x=a/b$, $y=c/d$, you'll get $2+x/y+y/x$. Now use that for any positive number the sum of that number and its reciprocal is at least $2$.
|
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|
Convergence/divergence of $\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}$ How do I show convergence/divergence of the series
$$\sum\limits_{n=1}^{\infty}\left(\cos\frac{1}{n}\right)^{n^3}?$$
I begin by writing $\left(\cos\frac{1}{n}\right)^{n^3} = e^{n^3\ln\left(\cos\frac{1}{n}\right)}$ and continue by Taylor expanding around $0$; first cosine, then ln. But I get nowhere. I would appreciate any help.
|
$\cos(\frac{1}{n}) = 1 - \frac{1}{2n^2} + o(\frac{1}{n^3})$. Let $a_n = \cos(\frac{1}{n})$ and $b_n = 1- \frac{1}{2n^2}$. We have $$\frac{a_n^{n^3}}{b_n^{n^3}} = (1 + \frac{o(\frac{1}{n^3})}{1-\frac{1}{2n^2}})^{n^3} = (1+c_n o(1))^{n^3} $$ where $c_n = \frac{1}{n^3} \cdot \frac{1}{1 - \frac{1}{2n^2}} = \frac{1}{n^3-\frac{n}{2}} $, so since $c_n \cdot n^3 \to 1$ we get $\frac{a_n^{n^3}}{b_n^{n^3}} \to 1$. So by asymptotics (note that $a_n,b_n$ are nonnetative for large $n$), the question is equivalent to convergence/divergence of $\sum_{n=1}^\infty (1-\frac{1}{2n^2})^{n^3}$. Now, taking $n'$th root we get $\exp(n^2\ln(1-\frac{1}{2n^2}))$. To find its limit, note that $\exp$ is continuous, so it is sufficient to find the limit of sequence $(n^2\ln(1-\frac{1}{2n^2}))$ which is equal to $-\frac{1}{2} \cdot \frac{\ln(1-\frac{1}{2n^2})}{-\frac{1}{2n^2}} \to -\frac{1}{2}$, so our limit tends to $\exp(-\frac{1}{2}) < 1$, and that means our series converges.
|
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"timestamp": "2023-03-29T00:00:00",
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|
A solution of Diophantine equation: $\big(x+y+z\big)^{3}=27x y z$ with $(x,y)∈Z$ A solution is:
$x=(r+s)^{3}(r-3s)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$y=8r^{3}(2s-r)^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
$z=(r^{2}-2r s+3s^{2})^{3}\big(r^{4}-4r^{3}s+4r^{2}s^{2}+3s^{4}\big)^{3}$
Are there any more?
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We have the identity:
$a^3+b^3+c^3=3abc$
Condition is, $(a+b+c)=0$
In "OP" solution if we take,
$a=(r+s)(r-3s)$
$b=2r(2s-r)$
$c=(r^2-2rs+3s^2)$
We get, $(a+b+c)=0$
Hence, $a^3+b^3+c^3=3abc$ ----(1)
Therefore by cubing both sides of equation (1),
$(a^3+b^3+c^3)^3=(3abc)^3=27(abc)^3 =27(a^3)(b^3)(c^3)$
Let, $(x,y,z)=(a^3,b^3,c^3)$
And we get:
$(x+y+z)^3=27xyz$
The constant, $w=(r^4-4r^3s+4r^2s^2+3s^4)$
is a multiplication factor & can be multiplied in if required.
|
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|
A parameterized log-sine integral equating to square of arctan I have a really convoluted proof of the following:
$$ (1) \quad \quad - \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } dy = \big(\arctan(\sqrt{r})\big)^2 $$
I proved it for $0<r<1.$ To my surprise it holds for complex $r$ if $-1<\Re{(r)}<1.$ Numerically it holds that
$$ - \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } dy = \big(\text{arccot}(\sqrt{r})\big)^2 $$
if $\Re{(r)}>1$ or $\Re{(r)}<-1.$ This is not surprising given that $\text{arccot}(x) =\arctan(1/x)$ for $x>0.$ However, there is a $\pi$ to be added if $x<0.$
Given the symmetry within the integral, (1) is sufficient, so long as the user makes sure that the $r$ put in on the right-hand side has a real part $-1<\Re{(r)}<1$
I would like a proof (or if the formula exists, a reference) of (1), and with special attention paid to the complexity of the parameter $r.$
|
From the generating function of the Chebyshev polynomials of the second kind:
\begin{equation}
\frac{1}{1-2xz+z^{2}}=\sum_{n=0}^{\infty}U_{n}\left(x\right)z^{n}
\end{equation}
valid for $\left|z\right|<1$, we chose $z=-r, x=\cos\left( 2y \right)$ to express
\begin{equation}
\frac{\sin(2 y) } {r + 1/r + 2 \cos{(2 y)} } =\sum_{n=0}^{\infty}(-1)^n \sin\left( 2(n+1)y \right)r^{n+1}
\end{equation}
as $U_n(\cos z)=\sin\left( (n+1)z \right)/\sin z$.
The Fourier series for $\ln\left( \sin(y/2) \right)$ reads
\begin{equation}
\log\sin\frac{y}{2}=-\log 2-\sum_{p=1}^\infty\frac{1}{p}\cos(py)\qquad\forall y\in(0,2\pi)
\end{equation}
and thus
\begin{align}
I&=- \int_0^\pi \frac{\sin(2 y) \log(\sin (y/2)) } {r + 1/r + 2 \cos{(2 y)} } \,dy\\
&=\int_0^\pi
\left( \log 2+\sum_{p= 1}^\infty\frac{1}{p}\cos(py) \right)\left( \sum_{n=0}^{\infty}(-1)^n \sin\left( 2(n+1)y \right)r^{n+1} \right)\,dy
\end{align}
Using symmetry properties, the contributions of the $\log2$ term and of the terms with even values of $p$ vanish:
\begin{align}
I&=\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\int_0^\pi\cos\left( (2s+1)y \right)\sin\left( 2(n+1)y \right)\,dy\\
&=\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\left[\frac{1}{2n-2s+1}+\frac{1}{2n+2s+3}\right]
\end{align}
A decomposition of the double sum can be written as
\begin{align}
I&=\sum_{n=0}^\infty\sum_{s=0}^n\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}+
\sum_{n=0}^\infty\sum_{s=n+1}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}+
\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n+2s+3}
\end{align}
By changing $k=s+n+1$ in the last double sum, we obtain
\begin{equation}
\sum_{n=0}^\infty\sum_{s=0}^\infty\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n+2s+3}=-\sum_{n=0}^\infty\sum_{k=n+1}^\infty\frac{(-1)^nr^{n+1}}{2n-2k+1}\frac{1}{2k+1}
\end{equation}
which is the opposite of the second double sum. Now, if $r^{1/2}$ is defined with its principal value, the integral can be expressed as
\begin{align}
I&=\sum_{n=0}^\infty\sum_{s=0}^n\frac{(-1)^nr^{n+1}}{2s+1}\frac{1}{2n-2s+1}\\
&=\sum_{n=0}^\infty\sum_{s=0}^n
\frac{(-1)^{s}r^{s+1/2}}{2s+1}
\frac{(-1)^{n-s}r^{n-s+1/2}}{2(n-s)+1}\\
&=\left( \sum_{t=0}^\infty
\frac{(-1)^{t}r^{t+1/2}}{2t+1} \right)^2\\
&=\left( \arctan\left(r^{1/2}\right) \right)^2
\end{align}
as proposed.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3704260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
How do I prove that $\max(x,\max(y,z)) = \max(\max(x,y),z))$ using an algebraic formula? The maximum of two numbers can be expressed by
$$\max(x,y) = \frac12\left(x+y+|x-y|\right)$$
Consequently, we can write
$$\max(x,\max(y,z))=\frac{1}{4}\left(2x+y+z+|y-z|+|2x-y-z-|y-z||\right)$$
$$\max(\max(x,y),z)=\frac{1}{4}\left(x+y+2z+|x-y|+|-x-y+2z-|x-y||\right)$$
Since we know the values on the left are equal, the expressions on the right should be, too. However, I don't know how to show that those expressions are the same.
|
From
$$\max(x,\max(y,z))=\frac{1}{4}\left(2x+y+z+|y-z|+|2x-y-z-|y-z||\right)$$
$$\max(\max(x,y),z)=\frac{1}{4}\left(x+y+2z+|x-y|+|-x-y+2z-|x-y||\right)$$
Let $y-z=a, x-y=b$, so $z=y-a, x=y+b, x-z=a+b$.
Then
$$\frac{1}{4}\left(x+2y+z+b+|a|+|2x-y-z-|y-z||\right)$$
$$\frac{1}{4}\left(x+2y+z-a+|b|+|-x-y+2z-|x-y||\right)$$
and
$$\frac{1}{4}\left(x+2y+z+b+|a|+|2x-y-z-|a||\right)$$
$$\frac{1}{4}\left(x+2y+z-a+|b|+|-x-y+2z-|b||\right)$$
and
$$\frac{1}{4}\left(x+2y+z+b+|a|+|a+2b-|a||\right)$$
$$\frac{1}{4}\left(x+2y+z-a+|b|+|-2a-b-|b||\right)$$
and using $|x|=x, x\gt0, |x|=-x, x\lt0$ we evaluate the last absolute term of the first and second equations independently for the four cases of the signs of $a$ and $b$
\begin{array}{|c|c|c|}
\hline
\frac{a}{b}&-&+\\
\hline
-&2|a+b|,2|a|&2|b|,2|a|\\
\hline
+&2|a+b|,2|a+b|&2|b|,2|a+b|\\
\hline
\end{array}
with the first entry in a cell being relevant to the first equation, etc..
For example if $a\gt0, b\lt0$ eqn 1 reads
$$|a+2b-|a||=|a+2b-a|=|2b|=2|b|$$
and eqn 2
$$|-2a-b-|b||=|-2a-b+b|=|-2a|=2|a|$$
as $-|b|=b$.
Replacing each last term with the associated cell, the equations match.
If $a,b$ are the same sign, $|a+b|=|a|+|b|$, otherwise they cancel as a whole.
For example $a\lt0,b\gt0$
$$b+|a|+|a+2b-|a||=b+|a|+2|a+b|$$
$$-a+|b|+|-2a-b-|b||=-a+|b|+2|a+b|$$
As $a\lt0, -a=|a|$ and as $b\gt0, |b|=b$
so we get
$$=b+|a|+2|a+b|$$
$$=|a|+b+2|a+b|$$
which are equal.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How to prove $\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)$? Recently I meet a problem ,it says
Suppose $a,b,c,x,y,z\in \mathbb{R}^+$,then
\begin{align*}
\frac{x}{y+z}(b+c)+\frac{y}{z+x}(a+c)+\frac{z}{x+y}(a+b)\geq
\sqrt{3(ab+bc+ca)}
\end{align*}
Fix $a,b,c$,then the original inequality is equal to
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \sqrt{3(ab+bc+ca)}+2(a+b+c)
\end{align*}
By using Cauchy's inequality,we can get
\begin{align*}
\frac{x+y+z}{y+z}(b+c)+\frac{x+y+z}{z+x}(a+c)+\frac{x+y+z}{x+y}(a+b)\geq \frac{1}{2}(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2
\end{align*}
So if we can proof (Since the original equality is true ,then the following equality must be true)
\begin{align*}
(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a})^2\geq 2\sqrt{3(ab+bc+ca)}+4(a+b+c)
\end{align*}
or
\begin{align*}
\sqrt{a+b}\sqrt{b+c}+\sqrt{b+c}\sqrt{c+a}+\sqrt{c+a}\sqrt{a+b}\geq \sqrt{3(ab+bc+ca)}+(a+b+c)\tag{*}
\end{align*}
then the problem is done.But I can't prove (*).
|
We'll prove a more general result:
If $x,y,z$ are the sides of a triangle, then
$$2\left(xy+yz+zx\right)\geq4S\sqrt3+x^2+y^2+z^2.$$
Indeed, as the $RHS$ is a decreasing function of $xyz,$ then according to the $uvw$ principles for triangle sides, we only need to check in the latter $y=z=1$ and $x\in[0,1]$ and get
$$x^2\left(x-1\right)^2\geq0,$$
which is obvious. We are done.
P.S. I just realised that I proved Finsler–Hadwiger theorem.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find an equivalent sequence as $n\to +\infty$ of $u_1>0, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}$ Let $u_1>0$ be a real number. Let us consider $(u_n)_{n\geq 1}$ the sequence such as:
$$
\forall n \geq 1, u_{n+1} = \frac{u_n}{n} + \frac{1}{n^2}\quad (\star)
$$
Find an equivalent of $u_n$ as $n\to +\infty$.
So I found a way to show that $u_n \sim \frac{1}{n^2}$, but I'm quite unhappy with this method because I feel like I found it by chance without understanding anything (I did a lot of trials and found this)
My method:
I showed by induction that $u_n \leq (u_1+1)$. Thus, $u_n\to 0$ considering $(\star)$.
Then, $nu_{n+1} = u_n + 1/n$. Thus (since $n+1 \sim n$), $nu_n \to 0$.
To end with, I have $n^2u_{n+1} = nu_n + 1$. Thus, $(n+1)^2 u_n \sim n^2u_n \to 1$.
How would you solve such a problem? Is there any more intuitive method that one may have done?
|
1. There is a heuristic argument which is useful for guessing the behavior of $u_n$: Rewrite the recurrence relation as
$$ u_{n+1} - u_n = \frac{u_n}{n} - u_n + \frac{1}{n^2}. $$
Its continuum analogue is the following differential equation:
$$ y' = \frac{y}{x} - y + \frac{1}{x^2}. $$
Using the standard method, this equation can be solved as:
$$ y(x) = x e^{-x} \int \frac{e^x}{x^3} \, \mathrm{d}x. $$
Then L'Hospital's Rule then tells that $y(x) \sim x^{-2}$ as $x \to \infty$. From this observation, we may as well expect that $u_n \sim n^{-2}$.
2. The above ansatz suggests that, in the recurrence relation for $u_{n+1}$, $\frac{1}{n^2}$ is the dominating term and $\frac{u_n}{n}$ is much smaller as $n\to\infty$. In particular, nesting this relation will produce an expansion with ever decreasing terms. This idea can be easily tested as follows:
Let $r_n = (n-1)u_n$. Then for $n \geq 2$,
$$ r_n = (n-1)u_{n} = \frac{1}{n-1} + \frac{r_{n-1}}{n-1}. $$
From this, we get
\begin{align*}
r_n
&= \frac{1}{n-1} + \frac{r_{n-1}}{n-1} \\
&= \frac{1}{n-1} + \frac{1}{(n-1)(n-2)} + \frac{r_{n-2}}{(n-1)(n-2)} \\
&= \frac{1}{n-1} + \frac{1}{(n-1)(n-2)} + \frac{1}{(n-1)(n-2)(n-3)} + \frac{r_{n-3}}{(n-1)(n-2)(n-3)} \\
&\qquad\vdots\\
&= \sum_{k=1}^{n-2} \frac{1}{(n-1)\cdots(n-k)} + \frac{r_2}{(n-1)!}
\end{align*}
Using this, it is not hard to conclude that $(n-1)^2 u_n \to 1$ as $n\to\infty$, and in fact, we can extract an asymptotic expansion of $u_n$ up to any prescribed order $\mathcal{O}(n^{-M})$.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Show $\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}$
I've been attempting to show that:
$$\sum_{k=1}^n {k+1\choose 2}{2n+1\choose n+k+1}={n\choose 1}4^{n-1}\\
\sum_{k=2}^n {k+2\choose 4}{2n+1\choose n+k+1}={n\choose 2}4^{n-2}$$
Can anyone give some direction?
|
Let's rearrange the sum as
$$
\eqalign{
& s(n) = \sum\limits_{k = 1}^n {\left( \matrix{
k + 1 \cr
2 \cr} \right)\left( \matrix{
2n + 1 \cr
n + k + 1 \cr} \right)} = \cr
& = \sum\limits_{k = \,0}^{n - 1} {\left( \matrix{
k + 2 \cr
2 \cr} \right)\left( \matrix{
2n + 1 \cr
n + k + 2 \cr} \right)} = \cr
& = \sum\limits_{0\, \le \,k\,} {\left( \matrix{
k + 2 \cr
2 \cr} \right)\left( \matrix{
2n + 1 \cr
n + k + 2 \cr} \right)} \cr}
$$
The summand is equal to
$$
\eqalign{
& t_{\,k} = \left( \matrix{ k + 2 \cr 2 \cr} \right)
\left( \matrix{ 2n + 1 \cr n + k + 2 \cr} \right) = \cr
& = {{\Gamma \left( {k + 3} \right)} \over {\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)}}
{{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( 3 \right)}} \cr}
$$
so that
$$
t_{\,0} = \left( \matrix{
2n + 1 \cr
n + 2 \cr} \right) = {{\Gamma \left( {2n + 2} \right)}
\over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}
$$
and the ratio is
$$
\eqalign{
& {{t_{\,k + 1} } \over {t_{\,k} }} = \cr
& = {{\Gamma \left( {k + 4} \right)}
\over {\Gamma \left( {k + 2} \right)\Gamma \left( {n + k + 4} \right)\Gamma \left( {n - 1 - k} \right)}}
{{\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)}
\over {\Gamma \left( {k + 3} \right)}} = \cr
& = {{\Gamma \left( {k + 4} \right)\Gamma \left( {k + 1} \right)\Gamma \left( {n + k + 3} \right)\Gamma \left( {n - k} \right)}
\over {\Gamma \left( {k + 3} \right)\Gamma \left( {k + 2} \right)\Gamma \left( {n + k + 4} \right)\Gamma \left( {n - 1 - k} \right)}} = \cr
& = {{\left( {k + 3} \right)\left( {n - 1 - k} \right)} \over {\left( {k + 1} \right)\left( {n + k + 3} \right)}}
= - {{\left( {k + 3} \right)\left( {k - \left( {n - 1} \right)} \right)} \over {\left( {k + 1} \right)\left( {k + n + 3} \right)}} \cr}
$$
That means that we can write $s(n))$ in terms of a Hypergeometric function
$$
s(n) = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}
{}_2F_{\,1} \left( {\left. {\matrix{
{\;3,\; - \left( {n - 1} \right)} \cr
{n + 3} \cr
} \;} \right|\; - 1} \right)
$$
Since
$$
1 + a - b = 1 + 3 + n - 1 = n + 3 = c
$$
we can apply [Kummer's theorem](https://en.wikipedia.org/wiki/Hypergeometric_function#Kummer's_theorem_(z_=_%E2%88%921)
and get
$$
\eqalign{
& s(n) = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}{}_2F_{\,1} \left( {\left. {\matrix{
{\;3,\; - \left( {n - 1} \right)} \cr
{n + 3} \cr
} \;} \right|\; - 1} \right) = \cr
& = {{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( {n + 3} \right)\Gamma \left( n \right)}}{{\Gamma \left( {n + 3} \right)\Gamma \left( {5/2} \right)}
\over {\Gamma \left( 4 \right)\Gamma \left( {3/2 + n} \right)}} = \cr
& = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)}}{{\Gamma \left( {2n + 2} \right)} \over {\Gamma \left( n \right)\Gamma \left( {3/2 + n} \right)}} \cr}
$$
Then by the duplication formula
$$
\Gamma \left( {2\,n + 2} \right) = 2^{\,2\,n + 1} {{\Gamma \left( {n + 1} \right)\Gamma \left( {n + 3/2} \right)} \over {\Gamma \left( {1/2} \right)}}
$$
we finally reach to
$$
\eqalign{
& s(n) = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)\Gamma \left( {1/2} \right)}}{{\Gamma \left( {n + 1} \right)\Gamma \left( {n + 3/2} \right)}
\over {\Gamma \left( n \right)\Gamma \left( {3/2 + n} \right)}}2^{\,2\,n + 1} = \cr
& = {{\Gamma \left( {5/2} \right)} \over {\Gamma \left( 4 \right)\Gamma \left( {1/2} \right)}}n\,2^{\,2\,n + 1} = {1 \over 8}n\,2^{\,2\,n + 1} = n\,2^{\,2\,n - 2} \cr}
$$
|
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"url": "https://math.stackexchange.com/questions/3706767",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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|
Triangles and medians Let G be the ABC barycenter. A line intersects the medians AD, BE and CF in X, Y and Z, respectively. Prove that
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}=3$$
By areas relations I found that
$\frac{XG\cdot YG}{BG\cdot DG}=\frac{6SG_{\triangle XYG}}{S_{\triangle ABC}}\\
\frac{XG\cdot ZG}{CG\cdot DG}=\frac{6SG_{\triangle XZG}}{S_{\triangle ABC}}\\
\frac{YG\cdot ZG}{BG\cdot CG}=\frac{3SG_{\triangle YZG}}{S_{\triangle ABC}}$
And using that
$S_{GYZ}=S_{GXZ}+S_{GXZ}$
I wrote
$$XG\cdot YG\cdot CG+XG\cdot ZG\cdot BG=2YG\cdot ZG\cdot DG$$
But I can't solve this problem.
Can someone help me?
Thanks for antetion!
|
Without loss of generality, we can scale and place the triangle so that its center of mass is in the origin and one of its vertices is on the $y$-axis at a unitary distance from the center of mass. So the vertices are $A(0,1)$, B$(p,q)$, and $C(-p,r)$. By well known properties of the medians, the midpoints of $BC$, $AC$, and $AB$ are, respectively:
$$D\Big(0, -\frac{1}{2}\Big)$$
$$E\Big(-\frac{p}{2}, -\frac{q}{2}\Big)$$
$$F\Big(\frac{p}{2}, -\frac{r}{2}\Big)$$
Also note that, because the $y$-coordinate of $D$ is the average of the $y$-coordinates of $B$ and $C$, we have $(q+r)/2=-1/2$ and then $q+r=-1$.
The medians correspond to the lines
$$AD\rightarrow\,\,\, x=0$$
$$BE\rightarrow\,\,\, y=\frac{q}{p}\,x$$
$$CF\rightarrow\,\,\, y=-\frac{r}{p}\,x $$
Now let us draw a line $y=mx+n$, not parallel to any of the medians. Its intersections with the medians $AD$, $BE$, and $CF$ are, respectively
$$X\left(0,n \right)$$
$$Y\left(\frac{np}{q-mp}, \frac{nq}{q-mp} \right)$$
$$Z\left(-\frac{np}{r+mp}, \frac{nr}{r+mp}\right)$$
Then we have
$$\frac{XD}{XG}=\left|\frac{n+1/2}{n}\right|=1+\frac{1}{2n}$$
$$\frac{YE}{YG}=\frac{ \sqrt{
\left( \frac{np}{q-mp}+\frac{p}{2}
\right)^2+\left(
\frac{nq}{q-mp}+\frac{q}{2} \right)^2 }}{ \sqrt{
\left( \frac{np}{q-mp}
\right)^2+\left(
\frac{nq}{q-mp}\right)^2 } }\\
=\sqrt{1+\frac{(np^2+nq^2)(q-mp) }{n^2(p^2+q^2)}+\frac{(p^2+q^2)(q-mp)^2}{4n^2(p^2+q^2)}
}\\
\sqrt{ 1+\frac{q-mp}{n}+ \frac{(q-mp)^2}{4n^2} }\\
= 1+\frac{q-mp}{2n} $$
$$\frac{ZF}{ZG}=\frac{ \sqrt{
\left( \frac{-np}{r+mp}-\frac{p}{2}
\right)^2+\left(
\frac{nr}{r+mp}+\frac{r}{2} \right)^2 }}{ \sqrt{
\left( \frac{-np}{r+mp}
\right)^2+\left(
\frac{nr}{r+mp}\right)^2 } }
\\
=\sqrt{1+\frac{(np^2+nr^2)(r+mp) }{n^2(p^2+r^2)}+\frac{(p^2+r^2)(r+mp)^2}{4n^2(p^2+r^2)}
}\\
\sqrt{ 1+\frac{r+mp}{n}+ \frac{(r+mp)^2}{4n^2} }\\
= 1+\frac{r+mp}{2n}
$$
Therefore
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}\\= 1+\frac{1}{2n}+1+\frac{q-mp}{2n} + 1+\frac{r+mp}{2n} \\
= 3+\frac{1}{2n}+\frac{q+r}{2n} $$
and since $q+r=-1$, we get
$$\frac{XD}{XG}+\frac{YE}{YG}+\frac{ZF}{ZG}= 3$$
|
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"url": "https://math.stackexchange.com/questions/3708717",
"timestamp": "2023-03-29T00:00:00",
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|
Evaluate: $\int \frac{dx}{(x^2+x+1)^2},$ without using any kind of substitutions. I know there exists a reduction formula for the integral: $$\int\frac1{(ax^2+bx+c)^n}\,dx.$$ But this uses substitutions. So for the integral in question, let:
\begin{align}
I&=\int\frac{dx}{(x^2+x+1)^2}\\\\
&=\int\frac{dx}{\{(x-\omega)(x-\omega^2)\}^2}\\\\
&=\int\frac{dx}{(x-\omega)^2(x-\omega^2)^2},
\end{align}
Where, $\omega$ is one of the complex cube roots of unity. According to me, resolving $\dfrac1{(x-\omega)^2(x-\omega^2)^2}$ into partial fractions, would serve our purpose. Is this a proper way to approach? Thanks in advance.
|
\begin{align}
I=\int \frac{dx}{(x^2+x+1)^2}&=\frac{1}{B}\int \frac{1}{2x+1}d(A-\frac{B}{x^2+x+1}) \\
&=\frac{1}{B}\left[ \frac{1}{2x+1}(A-\frac{B}{x^2+x+1})+\int (A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2}dx \right] \\
\end{align}
Note
\begin{align}
&(A-\frac{B}{x^2+x+1})\frac{2}{(2x+1)^2} \\
&=\frac{Ax^2+Ax+A-B}{x^2+x+1} \frac{2}{4x^2+4x+1}
\end{align}
So,we can let
$$ A=4,A-B=1\Rightarrow A=4,B=3 $$
Thus
\begin{align}
I&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+2\int \frac{1}{x^2+x+1} dx \right] \\
&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+2\int \frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}} dx \right] \\
&=\frac{1}{3}\left[ \frac{2x+1}{x^2+x+1}+\frac{4}{\sqrt{3}}\arctan \frac{2x+1}{\sqrt{3}} \right]+C \\
\end{align}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverse trigonometric equation and finding value of given expression If $\arctan(4) = 4 \arctan(x)$ then value of $x^5-7x^3+5x^2+2x+9870$ is?
I used $2\arctan(x) = \arctan(2x/1-x^2)$ twice for RHS of the equation which gave me $x^4+x^3-6x^2-x+1=0$ and I am clueless how to proceed after that.
Also I am not sure about using the formula I mentioned as its valid only when $|x|<1$.
Please help me in this regard
|
Hint
From where you have left, use
$$x^5-7x^3+5x^2+2x=x(x^4+x^3-6x^2-x+1)-(x^4+x^3-6x^2-x+1)+1$$
|
{
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"url": "https://math.stackexchange.com/questions/3718407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$? If $abc=1$, then how do you prove $\frac{b-1}{bc+1}+\frac{c-1}{ac+1}+\frac{a-1}{ab+1} \geq 0$?
I tried substitution on the bottom (for example $\frac{b-1}{\frac{1}{a}+1}$), but I then a very similar term. What should I do?
I can not find a way to decompose of combine.
Also, how do you prove that $x^4+y^4+z^2 \ge xyz \sqrt{8}$?
I got $x^4+y^4+z^2\geq3\sqrt[3]{x^4*y^4*z^2}=3xy\sqrt[3]{xyz^2}.$
I do not know what to do from here.
|
Another way for the first inequality.
For positives $x$, $y$ and $z$ we need to prove that:
$$\sum_{cyc}(x^3-x^2z)\geq0.$$
Indeed,
$$\sum_{cyc}(x^3-x^2z)=\sum_{cyc}(x^3-xy^2)=$$
$$=\sum_{cyc}x(x+y)(x-y)=\sum_{cyc}\left((x^2+x)(x-y)-\frac{2}{3}(x^3-y^3)\right)=$$
$$=\frac{1}{3}\sum_{cyc}(x-y)(3x^2+3xy-2x^2-2xy-2y^2)=\frac{1}{3}\sum_{cyc}(x-y)^2(x+2y)\geq0.$$
|
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|
Find the maximum of $P$ Given $a, b, c \geq 0$ and $a + b + c = 6$. Find the maximum of:
$$P = (a-b)(b-c)(c-a)$$
My guess is the maximum of $P$ would be just $0$ but I don't know whether this guess is true
And if it's true, how do I prove it?
|
Suppose $c = \min\{a,b,c\},$ so
$$(a-c)^2 \leqslant a^2, \quad (b-c)^2 \leqslant b^2.$$
By the AM-GM inequality we have
$$\begin{aligned}P^2 = (a-b)^2(b-c)^2(c-a)^2 & \leqslant a^2b^2(a-b)^2 = \frac14 \cdot 2ab \cdot 2ab \cdot (a-b)^2 \\& \leqslant \frac14\left(\frac{2ab+2ab+(a-b)^2}{3}\right) ^3\\&=\frac{(a+b)^6}{108}\leqslant \frac{(a+b+c)^6}{108}\end{aligned}$$
Therefore
$$(a+b+c)^3 \geqslant 6 \sqrt 3 (a-b)(b-c)(c-a).$$
Equality occur when $a^2+b^2=4ab,\,c= 0$ and permutation.
Because $a+b+c=6,$ so
$$(a-b)(b-c)(c-a) \leqslant \frac{(a+b+c)^3}{6\sqrt 3} = 12\sqrt 3.$$
Equality occur when
$$a^2+b^2=4ab,\,c= 0,\,a+b+c=6. \quad (1)$$
Solve $(1)$ we get $a = 3+\sqrt3,\,b = 3-\sqrt3,\, c = 0$ or $a = 3-\sqrt3, \,b = 3+\sqrt3,\, c = 0.$
Thefore $P_{\max} = 12\sqrt 3.$
|
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|
How to evaluate this integral $\int \frac{dx}{16+5^{2-3x}}$? Evaluate the following integral
$$\int \frac{dx}{16+5^{2-3x}}=?$$
My attempt:
Assume that $2-3x=t$, $-3dx=dt$, $dx=-\frac{dt}{3}$
$$\int \frac{-dt/3}{16+5^{t}}=-\frac13\int \frac{dt}{16+5^{t}}$$
let $ 5^t=u$, $\implies 5^t\ln 5\ dt=du$, $dt=\frac{du}{u\ln 5}$
$$-\frac13\int \frac{\frac{du}{u\ln 5}}{16+u}$$
$$=-\frac1{3\ln 5}\int \frac{du}{u(16+u)}$$
$$=-\frac1{3\ln 5}\int \frac{1}{16}(\frac{1}{u}-\frac{1}{16+u})du$$
$$=-\frac1{48\ln 5}(\ln u-\ln(u+16))$$
I back substituted, $u=5^t$
$$=-\frac1{48\ln 5}(\ln 5^t-\ln(5^t+16))$$
I again back substituted, $t=2-3x$
$$=-\frac1{48\ln 5}(\ln 5^{2-3x}-\ln(5^{2-3x}+16))$$
after simplification i got the answer
$$=\frac1{48\ln 5}\ln\left(\frac{16+5^{2-3x}}{5^{2-3x}}\right)+c$$
In this integral I used two succesive substitutions, which is a bit lengthier. Can the same integral be calculated by some easier method? Can I evaluate by a single substitution? some body please help me. Thanks
|
Yes, you can use single substitution
Let $16+5^{2-3x}=t\implies 5^{2-3x}\ln 5\cdot (-3)=dt$, $dx=\frac{dt}{3(16-t)\ln 5}$
$$\int \frac{dx}{16+5^{2-3x}}=\int \frac{1}{t}\frac{dt}{3(16-t)\ln 5}$$
$$=\frac{1}{3\ln5}\int \frac{dt}{t(16-t)}$$
$$=\frac{1}{3\ln5}\frac{1}{16}\int\left( \frac{1}{t}-\frac{1}{t-16}\right)\ dt$$
$$=\frac{1}{48\ln5}\int \left(\frac{1}{t}-\frac{1}{t-16}\right)\ dt$$
|
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|
Evaluate $1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots$ My attempt
$$1-\frac{1}{3\cdot 3}+\frac{1}{5\cdot 3^2}-\frac{1}{7\cdot 3^3}+\cdots=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}$$
By Leibniz alternative test for convergence. It is a convergent alternative series. How do I evaluate this limit?
|
$$
\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)3^{n-1}}=\sum_{n=0}^\infty \frac{(-1)^{n}}{(2n+1)3^{n}}=\sqrt{3}\sum_{n=0}^\infty \frac{(-1)^{n}(1/\sqrt{3})^{2n+1}}{(2n+1)}
$$
$$
=\sqrt{3}\arctan(1/\sqrt{3}) = \frac{\sqrt{3}\pi}{6}
$$
|
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|
Prove $x^4 + x^2 +1$ is always greater than $x^3 + x$ Let's say P is equal to $x^4 + x^2 +1$ and $Q$ is equal to $x^3 + x$.
For $x <0$, $P$ is positive and $Q$ is negative. Hence, in this region, $P>Q$.
For $x=0$, $P>Q$.
Also, for $x = 1$, $P>Q$.
For $x > 1$, I factored out $P$ as $x^2(x^2+1) + 1$ and $Q$ as $x(x^2+1)$. For $x > 1$, $x^2(x^2+1) > x(x^2+1)$, hence $P>Q$.
The part where I have the problem is I can't prove this for the range $0 < x < 1$ without the help of a graphing calculator. Can anyone help?
What I've done in this region so far is:
*
*Prove that $P$ and $Q$ is always increasing in this region,
*The range for $P$ starts from $1 < P < 3$, and
*The range for $Q$ starts from $0 < Q < 2$.
The only thing I need to prove now is that $P$ and $Q$ will not intersect at $0 < x < 1$, but I can't prove this part.
|
For $x\leq0$ it's obvious.
But for $x>0$ we obtain:
$$x^4+x^2+1-(x^3+x)=x^4-2x^3+3x^2-2x+1+x^3+2x^2+x=$$
$$=(x^2-x+1)^2+x(x+1)^2>0.$$
|
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|
Consider $\dot{x}=4x^{2}-16$. I am solving the ODE above, it is a question from Strogatz Nonlinear dynamics and chaos, chapter 2 question 2.2.1.
Question
\begin{equation}
\dot{x}=4x^{2}-16
\end{equation}
Answer
\begin{equation}
\frac{\dot{x}}{4x^{2}-16} = 1\\
\frac{\dot{x}}{x^{2}-4} = 4\\
{{dx\over dt}\over x^2-4}=4 \\
{{dx\over dt}\over x^2-4}. dt=4. dt \\
{dx\over x^2-4}=4dt \\
\int \frac{1}{x^{2}-4} dx = \int 4 dt \\
\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} \\
x = 2 \frac{1 + C_{2}e^{16t}}{1 - C_{2}e^{16t}}
\end{equation}
\begin{equation}
C_{2}(t=0) = \frac{x-2}{x+2}
\end{equation}
Summary
I am looking to understand the intermediary step in the proof above. How do we get to this step $\frac{1}{4} \ln(\frac{x-2}{x+2}) = 4t + C_{1} $ from the previous step. Can we remove the constant $\frac{1}{4} $ then integrate the remaining portion?
|
The author has used the method of "partial fractions" to write
$$
\frac{1}{x^2-4} = \frac{A}{x-2} + \frac{B}{x+2},
$$
(solving for $A$ and $B$, which I'm not going to do), integrated each of the right hand items to get a "log" term, and then combined a difference-of-logs into a log-of-a-quotient. There's nothing subtle here except skipping steps from calculus class.
|
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|
How can I solve this definite integral: $\int_{0}^{a}\frac{x^4dx}{\sqrt{a^2-x^2}}$
Evaluate $$\int_{0}^{a}\dfrac{x^4dx}{\sqrt{a^2-x^2}}$$
I tried taking $t$ as
$$t = \sqrt{a^2-x^2}$$
Thus my final integral became
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt$$
but I couldn't go any further in solving this integral.
I also tried by taking $t$ as
$$t = a\sin^{-1}{x}$$
But I don't know how to solve the resulting integrand.
Also, can the king's rule be applied here? If yes then how?
|
You can proceed as follows
Let $t=a\sin\theta\implies dt=a\cos\theta d\theta$
$$\int_{0}^{a}(a^2-t^2)^{3/2}dt=\int_{0}^{\pi/2}(a^2-a^2\sin^2\theta)^{3/2}a\cos\theta \ d\theta$$
$$=\int_{0}^{\pi/2}(a^{3}\cos^3\theta) a\cos\theta \ d\theta$$
$$=a^4\int_{0}^{\pi/2}\cos^4\theta d\theta$$
Using: $\color{blue}{\int_0^{\pi/2}\sin^m\theta\cos^n\theta\ d\theta=\dfrac{\Gamma(\frac{m+1}{2})\Gamma(\frac{n+1}{2})}{2\Gamma(\frac{m+n+2}{2})}}$ ,
$$=a^4\frac{\Gamma(\frac{4+1}{2})\Gamma(\frac{0+1}{2})}{2\Gamma(\frac{4+0+2}{2})}$$
$$=a^4\frac{\frac32\frac12\Gamma(\frac{1}{2})\Gamma(\frac{1}{2})}{2\cdot 2}$$
$$=a^4\frac{\frac34\sqrt{\pi}\cdot \sqrt{\pi}}{2\cdot 2}=\frac{3\pi}{16}a^4$$
|
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|
Inequality 6 deg For $a,b,c\ge 0$ Prove that $$4(a^2+b^2+c^2)^3\ge 3(a^3+b^3+c^3+3abc)^2$$
My attempt: $$LHS-RHS=12(a-b)^2(b-c)^2(c-a)^2+2(ab+bc+ca)\sum_{sym} a^2(a-b)(a-c)$$
$$+\left(\sum_{sym} a(a-b)(a-c)\right)^2+14\left(\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\right)+\sum_{sym} c^2(a-b)^2[2(a+b)^2+(a-b)^2]$$
Since $\sum_{sym} x^3+3xyz-\sum_{sym} xy(x+y)\ge 0$, set $(x,y,z)\rightarrow (ab,bc,ca)$ we obtain $$\sum_{sym} a^3b^3+3a^2b^2c^2-abc\sum_{sym} ab(a+b)\ge 0$$
Thus $LHS-RHS\ge 0$
I think this's a complicated solution and hard to find it by hand :">
Please give me a simplier solution. Thank you very much.
|
By the Cauchy-Schwarz inequality we have
$$(a^3+b^3+c^3+3abc)^2 = \left[\sum a(a^2+bc)\right] \leqslant \sum a^2 \sum (a^2+bc)^2.$$
Therefore we will show that
$$4(a^2+b^2+c^2)^2 \geqslant 3[(a^2+bc)^2+(b^2+ca)^2+(c^2+ab)^2],$$
equivalent to
$$a^4+b^4+c^4 + 5(a^2b^2+b^2c^2+c^2a^2) \geqslant 6abc(a+b+c). \quad (1)$$
Which is true because
$$a^4+b^4+c^4 \geqslant a^2b^2+b^2c^2+c^2a^2 \geqslant abc(a+b+c).$$
Note. The sum of squares of $(1)$
$$(a^2+b^2+c^2+ab+bc+ca)(a^2+b^2+c^2-ab-bc-ca)+2 \sum a^2(b-c)^2 \geqslant 0.$$
Zhaobin have posted it before
|
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|
Solving trigonometric equations: $\cos(3x)+ \cos(x)=0$ Recently I worked on a problem where I had to solve $$\cos(3x)+\cos(x)=0$$
When I tried calculating it by evaluating $x$, $x=2nπ+\dfrac{\pi}{4}$, $x=2n\pi-\dfrac{\pi}{4}$, or $x=2n\pi+\dfrac{\pi}{2}$ was the solution I reached.
Unfortunately, it was apparently wrong. Is there another way to solve this problem?
|
Use the cosine of sums formula to decompose $\cos 3x$.
$$\cos 3x = \cos (2x + x) = \cos 2x \cdot \cos x - \sin x \cdot \sin 2x$$
The identities $\cos 2x = 2 \cos^2 x - 1$ and $\sin 2x = 2 \sin x \cos x$ are widely known. These are also derived from the cosine of sums and sine of sums formulas respectively. Plugging them into our last result, we have
\begin{align}
\cos 3x &= \left(2 \cos^2 x - 1\right) \cos x - \sin x \cdot 2 \sin x \cos x \\
&= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\
\end{align}
Since $\sin^2 x = 1 - \cos^2 x$, we have
\begin{align}
\cos 3x &= 2\cos^3 x - \cos x - 2\sin^2 x \cdot \cos x \\
&= 2\cos^3 x - \cos x - 2 \left(1 - \cos^2 x\right) \cos x \\
&= 2\cos^3 x - \cos x - 2 \cos x + 2 \cos^3 x \\
&= 4\cos^3 x - 3 \cos x
\end{align}
At this point, we have eliminated the $\sin$'s and $\cos 3x$ is expressed entirely in terms of a polynomial of $\cos x$. Plugging this into our equation $\cos^3 x + \cos x = 0$, we have
\begin{align}
\left(4\cos^3 x - 3 \cos x\right) + \cos x &= 0 \\
4\cos^3 x - 2 \cos x &= 0 \\
2\cos^3 x - \cos x &= 0
\end{align}
Taking $y = \cos x$, this reduces to the polynomial
$$2y^3 - y = 0 \Rightarrow y(2y^2 - 1) = 0 \Rightarrow y(y - \sqrt{2}/2)(y + \sqrt{2}/2) = 0$$
which gives three distict roots $y = 0$, $y = -\sqrt{2}/2$ and $y = \sqrt{2}/2$. Then, we plug-in $y = \cos x$ into each and solve them separately.
For $y = 0$, we have $\cos x = 0$. This holds for $x = \pi/2$ plus any integer multiple of $\pi$. These are exactly the points where $\cos x$ intersects the $x$-axis.
For $y = -\sqrt{2}/2$, we have $\cos x = -\sqrt{2}/2$. From here, we have exactly the bottom points of the cosine curve $\pm \pi/4$.
For $y = +\sqrt{2}/2$, we have $\cos x = +\sqrt{2}/2$. From here, we have exactly the peak points of the cosine curve $\pm \pi/4$.
Taking the union of these solution sets, we get the solution set for $\cos^3 x + \cos x = 0$ which are exactly the multiples of $\pi/4$ that are not multiples of $\pi/2$.
|
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|
Solve the equation $\operatorname{arcsinh}=\operatorname{arcsech}(x)$ analytically I am trying to obtain an analytical solution of the equation.
$$\operatorname{arcsinh}(x) = \operatorname{arcsech}(x)$$
Equating the logarithmic definitions leads to the rather unwieldy equation
$$x^4+x^3\sqrt{x^2+1} +x^2 -1.0 -\sqrt{1-x^2}$$
Needless to say I am struggling to obtain an expression for x ! Can anyone offer a solution ?
|
$$\log(x+\sqrt{x^2+1})=\log\left(\frac1x+\sqrt{\frac1{x^2}-1}\right)$$
is equivalent to
$$x^2-1=\sqrt{1-x^2}-x\sqrt{x^2+1}.$$
Then with squaring,
$$x^4-2x^2+1=1-x^2-2x\sqrt{1-x^4}+x^2(x^2+1)$$
simplifies to
$$x=0\lor x=\sqrt{1-x^4}.$$
The last equation can be reduced to biquadratic.
$$x=\dfrac1{\sqrt\phi}.$$
|
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|
The quotient of a quotient group by another quotient group Let $G$, $M$, and $N$ be the cyclic groups given by
$$
G \colon= \left\langle a \colon a^{12} = e \right\rangle = \left\{ e = a^0, a, a^2, \ldots, a^{11} \right\},
$$
$$
M \colon= \left\langle a^2 \right\rangle = \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\},
$$
and
$$
N \colon= \left\langle a^6 \right\rangle = \left\{ e, a^6 \right\}.
$$
Then of course $M$ is a normal subgroup of $G$, and of course $N$ is a normal subgroup of both $G$ and $M$.
Thus the quotient groups $G/M$, $G/N$, and $M/N$ are well-defined.
In fact, we have
$$
\begin{align}
G/M &= \left\{ M, aM, a^3M, a^5M, a^7M, a^9M, a^{11}M \right\} \\
&= \left\{ M, aM \right\} \\
&= \left\{ \, \left\{ e, a^2, a^4, a^6, a^8, a^{10} \right\}, \, \left\{ a, a^3, a^5, a^7, a^9, a^{11} \right\} \, \right\},
\end{align}
$$
$$
\begin{align}
G/N &= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N, a^7N, a^8N, a^9N, a^{10}N, a^{11}N \right\} \\
&= \left\{ N, aN, a^2N, a^3N, a^4N, a^5N \right\} \\
&= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^4, a^{10} \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\},
\end{align}
$$
and
$$
\begin{align}
M/N &= \left\{ N, a^2N, a^4N, a^8 N, a^{10}N \right\} \\
&= \left\{ N, a^2N, a^4 N \right\} \\
&= \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}.
\end{align}
$$
Furthermore, as $M$ is a normal subgroup of $G$, so the quotient group $M/N$ is a normal subgroup of the quotient group $G/N$.
Therefore we can consider the quotient group $(G/N)/(M/N)$.
Now my question is, is the following construction valid?
We first note that
$$
\begin{align}
\left\{ a, a^7 \right\} (M/N) &= \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}.
\end{align}
$$
And also
$$
\begin{align}
\left\{ a^3, a^9 \right\} (M/N) &= \left\{ a^3, a^9 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a^3, a^9 \right\} \left\{ e, a^6 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a^3, a^9 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \\
&= \left\{ \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\}, \, \left\{ a, a^7 \right\} \, \right\} \\
&= \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\}.
\end{align}
$$
Thus we have shown that
$$
\left\{ a, a^7 \right\} (M/N) = \left\{ a^3, a^9 \right\} (M/N).
$$
Similarly we can show that
$$
\left\{ a, a^7 \right\} (M/N) = \left\{ a^5, a^{11} \right\} (M/N).
$$
Using the calculations above, we find that
$$
\begin{align}
(G/N)/(M/N) &= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) , \ \left\{ a^3, a^9 \right\}(M/N), \ \left\{ a^5, a^{11} \right\} (M/N) \ \right\} \\
&= \left\{ \ M/N, \ \left\{ a, a^7 \right\} (M/N) \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ a, a^7 \right\} \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\} \left\{ e, a^6 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^2, a^8 \right\}, \, \left\{ a, a^7 \right\} \left\{ a^4, a^{10} \right\} \, \right\} \ \right\} \\
&= \left\{ \ \left\{ \, \left\{ e, a^6 \right\}, \, \left\{ a^2, a^8 \right\}, \, \left\{ a^4, a^{10} \right\} \, \right\}, \ \left\{ \, \left\{ a, a^7 \right\}, \, \left\{ a^3, a^9 \right\}, \, \left\{ a^5, a^{11} \right\} \, \right\} \ \right\}.
\end{align}
$$
Of course the quotient group $(G/N)/(M/N)$ is isomorphic to the quotient group $G/M$.
Is my construction correct?
Is each and every detail of my calculation above correct? Or, have I made any errors or logical / mathematical mistakes?
Last but not the leat, is my typesetting logically correct and clear enough as well? Or, is there a better way of presenting the work above?
|
Not only is your conclusion correct, you can probably find it stated as a theorem (possibly with general proof) in your favorite group-theory textbook [e.g it is in my copy of Fraleigh 3rd edition].
|
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|
Added angle formula to solve this indefinite integral $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ Starting from this very nice question Integrate $\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx$ and the relative answers, I would to understand because this integral $$\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx \tag 1$$
must be split thus:
$$\int \frac{2\cos{x}-\sin{x}}{3\sin{x}+5\cos{x}} \; dx=\color{red}{\int A\left(\frac{3\sin{x}+5\cos{x}}{3\sin{x}+5\cos{x}}\right) +B \left(\frac{ 3\cos{x}-5\sin{x}}{3\sin{x}+5\cos{x}}\right)\; dx}$$
or it can be splitted in a different way.
Using the added angle formula (for numerator and denominator of the $(1)$) $$a\sin x+b\cos x=\lambda \sin (x+\phi)$$ if $\lambda=\sqrt{a^2+b^2}$ and $\tan \phi=b/a \ $ or $$a\sin x+b\cos x=\lambda \cos (x+\varphi)$$ with $\tan \varphi=-a/b \ $ is it possible to obtain the same result?
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$I=\int\frac{2\cos x-\sin x}{3\sin x+5\cos x }\,dx $
$2\cos x-\sin x=\sqrt5 \cos(x+a) $, where $\tan a=1/2$ and $ 3\sin x+5\cos x=\sqrt{34}\cos(x-b) $, where $\tan b=3/5$
$I=\int \frac{\sqrt 5 \cos(x+a) } {\sqrt{34}\cos(x-b)} \, dx$ Substitute $t=x-b$ so that
$I=\sqrt{5/34}\int \frac{\cos(t+a+b) } {\cos t}\, dt$
The integrand is now: $\cos(a+b) - \tan t\sin(a+b) $ Can you take it from here?
|
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|
Let $\langle x_n\rangle$ be a recursive relation. Find $\lim_{n\to\infty}\frac {x_n}{n^2}.$ Let $\langle x_n\rangle$ be a recursive relation given by $$x_{n+1}=x_n+a+\sqrt {b^2+4ax_n}, n\geq0, x_0 =0$$ and $a$ and $b$ are fixed positive integers. Find $$\lim_{n\to\infty}\frac {x_n}{n^2}.$$
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It is practical to set $x_n=a z_n$ and $r=\frac{b}{a}$ to get rid of a parameter:
$$ z_{n+1} = z_{n} + 1 + \sqrt{4z_n+r^2},\qquad z_0=0 $$
Obviously $z_n\geq n$, since $z_{n+1}\geq z_{n}+1$, but also
$$ z_{n+1} \geq (\sqrt{z_n}+1)^2, $$
hence by setting $z_{n}=w_n^2$ we get $w_{n+1}\geq w_n+1$ and $z_n\geq n^2$. On the other hand
$$ z_{n+1}-(\sqrt{z_n}+1)^2 = \sqrt{4z_n+r^2}-\sqrt{4z_n} = \frac{r^2}{\sqrt{4z_n}+\sqrt{4z_n+r^2}}\leq \frac{r^2}{4n} $$
$$ w_{n+1}^2 - (w_n+1)^2 \leq \frac{r^2}{4n} $$
$$ w_{n+1}-(w_n+1) \leq \frac{r^2}{4n(w_{n+1}+w_n+1)} \leq \frac{r^2}{8n(n+1)} $$
hence $w_n=n+O(1)$, $z_n=n^2+O(n)$ and $x_n = \color{red}{a}n^2+O(n)$.
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|
If $ 3a+2b+c=7$ then find minimum value of $ a^2+b^2+c^2$ Question:- If $ 3a+2b+c=7$ then find the minimum value of $ a^2+b^2+c^2 $.
I used vectors to solve this problem.
Let $$α=3\hat{i}+2\hat{j}+\hat{k}$$
$$β=a\hat{i}+b\hat{j}+c\hat{k}$$
Using Cauchy-Schwarz inequality
we have, $|α.β|\le |α| |β|$
$=|3a+2b+c|\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
$= 7\le\sqrt{14}\sqrt{a^2+b^2+c^2}$
So, $a^2+b^2+c^2\ge \frac72$
Therefore, the minimum value of $a^2+b^2+c^2$ is $\frac72$
I want to know are there any other method to find the minimum value of
$a^2+b^2+c^2$ such as using inequalities and calculus by assuming function $f(x,y,z)=x^2+y^2+z^2$.
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Let $P(a, b, c)$ be any point on $(\pi): 3 x+2 y+z=7$.$\\\\$
$\\ OP^{2}=a^{2}+b^{2}+c^{2}$ is the minimum iff OP $\perp(\pi) \Leftrightarrow \displaystyle O P =\left|\frac{3(0)+2(0)+7}{\sqrt{3^{2}+2^{2}+1^{2}}}\right| =\frac{7}{\sqrt{14}} =\sqrt{\frac{7}{2}}$
$\therefore$ the minimun value of $\displaystyle a^{2}+b^{2}+c^{2} \textrm{ is }\frac{7}{2}.$
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|
How to evaluate this integral? $\int^1_0\frac{(\frac{1}{2}-x)\ln (1-x)}{x^2-x+1}\mathrm{d}x$ $$\int^1_0\frac{\left(\frac{1}{2}-x\right)\ln (1-x)}{x^2-x+1}\mathrm{d}x$$
The indefinite integral didn't seem to be helpful...which include the $\operatorname{Li}(x)$.
Also, I can't set up a parameter to differentiate it.
Any hint will be helpful. Thanks!
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First, substitute $y=1-x$:
$$
\Rightarrow \int _0^1 \frac{(y-1/2)\log(y)}{y^2-y+1}\,dy
$$Then use integration by parts with $u=\log(y)$:
$$
=\left. \frac{1}{2}\log(y^2-y+1)\log(y)\right|_{0}^{1} -\frac{1}{2}\int_0^1\frac{\log(y^2-y+1)}{y}\,dy
$$The boundary terms vanish (check this yourself). Note that $y^3+1=(y^2-y+1)(y+1)$, so we have
$$
-\frac{1}{2}\int_0^1\frac{\log(y^2-y+1)}{y}\,dy = -\frac{1}{2}\int_0^1\frac{\log((y^3+1)/(y+1))}{y}\,dy
$$
$$
= \frac{1}{2}\int_0^1\frac{\log(y+1)}{y}\,dy- \frac{1}{2}\int_0^1\frac{\log(y^3+1)}{y}\,dy
$$Now invoke the Maclaurin series for $\log(1+z)$:
$$
= \frac{1}{2} \int _0^1 \sum_{k=0}^{\infty} \frac{(-1)^k y^k}{k+1}\,dy-\frac{1}{2} \int _0^1 \sum_{k=0}^{\infty} \frac{(-1)^k y^{3k+2}}{k+1}\,dy
$$Switch the integral and the sum and use the power rule:
$$
= \frac{1}{2} \sum_{k=0}^{\infty} \int _0^1 \frac{(-1)^k y^k}{k+1}\,dy-\frac{1}{2} \sum_{k=0}^{\infty}\int _0^1 \frac{(-1)^k y^{3k+2}}{k+1}\,dy
$$
$$
= \frac{1}{2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}-\frac{1}{2\cdot 3} \sum_{k=0}^{\infty} \frac{(-1)^k }{(k+1)^2}
$$
$$
= \frac{1}{3} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2}
$$If you buy that $\sum_{k=0}^{\infty} \frac{1}{(k+1)^2}=\frac{\pi^2}{6}$, you can convince yourself the alternating version sums to $\frac{\pi^2}{12}$. This gives the total of $\frac{\pi^2}{36}$.
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|
Calculate $\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$ $$\int_{1}^{\phi}\frac{x^{2}+1}{x^{4}-x^{2}+1}\ln\left(x+1-\frac{1}{x}\right) \mathrm{dx}$$
Insane integral! So far I have tried to complete the square for the denominator then substitute and use taylor series for the natural logarithm about x=0. Is this integral possible?!
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I assume that $\phi$ is the golden ratio. Consider $u=x-\frac{1}{x}$ so that $\frac{x^2}{x^2+1} du=dx$:
$$I=\int_0^1 \frac{x^2}{x^2+1} \cdot \frac{x^2+1}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{x^2}{x^4-x^2+1} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{1}{x^2-1+\frac{1}{x^2}} \ln{\left(1+u\right)} \; du$$
$$I=\int_0^1 \frac{1}{u^2+1} \ln{\left(1+u\right)} \; du$$
Now, let $u=\tan{t}$:
$$I=\int_0^{\frac{\pi}{4}} \ln{\left(1+\tan{t}\right)} \; dt$$
Then, substitute $w=\frac{\pi}{4}-t$
$$I=\int_0^{\frac{\pi}{4}} \ln{\left(1+\tan{\left(\frac{\pi}{4}-w\right)}\right)} \; dw$$
Use the tangent angle addition formula:
$$I=\int_0^{\frac{\pi}{4}} \ln{\left(1+\frac{1-\tan{w}}{1+\tan{w}}\right)} \; dw$$
$$I=\int_0^{\frac{\pi}{4}} \ln{2} \; dw-\int_0^{\frac{\pi}{4}} \ln{\left(1+\tan{w}\right)} \; dw$$
Remember that the second integral is $I$:
$$2I=\int_0^{\frac{\pi}{4}} \ln{2} \; dw$$
$$I= \frac{\pi \ln{2}}{8}$$
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|
Sum of geometric series when exponent is $2n$, not $n$? I have a probability below which denotes the chance of catching a fish.
$$P = \left(\frac{1}{4}\right) + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^2 + \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^4 + \dots $$
I can find a generalized form of $P$ by assuming the first term is $\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^0$ to come up with an infinite series:
$$P = \sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}$$
I'm learning about series now and I know that if $|r| \geq 1$ the series is divergent. Here I have $|r| = \dfrac{3}{4}$ so there is a sum. I also have $a = \dfrac{1}{4}$. In the examples, the series have an exponent of $n$ but none have $2n$. Were it to be $n$ instead, I could find the sum as:
$$P = \frac{a}{1-r} = \frac{\frac{1}{4}}{1-\frac{3}{4}} = 1$$
But this is with $n$, not $2n$, where the answer I seek is $\dfrac{4}{7}$. How can I approach this number? How do I reconcile the $2n$ and is there a formula for this like there was when it was $n$? I can do partial sums and find an approximation but what is the sum as $n \to \infty$?
I've marked the question calculus as this is a unit in Stewart's Early Transcendentals calculus text.
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$$\sum_{n=0}^\infty \left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{2n}=\frac{1}{4}\sum_{n=0}^\infty\left(\frac{9}{16}\right)^{n}=\frac{1}{4} \frac{1}{1- \frac{9}{16}}=\frac{4}{7}.$$
|
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Find coordinates of point in $\mathbb{R^4}$. There is a problem here in Q. $5$ on the last page. It states to find coordinates of point $p$.
Taking point $a=(3,2,5,1), \ b=(3,4,7,1), \ c= (5,8,9,3)$.
Also, $b$ has two coordinates in common with $a$, and $p$ lies on the same line as $a,b$.
So, those two coordinates of $p$ are same as $a,b$. Hence, $p= (3,x,y,1)$; where $x,y\in \mathbb{R}$ are unknown.
Given that $\triangle acp, \triangle bcp$ are right-angled; get:
$1. \ \ \triangle acp:\ \ \ \ \ {ac}^2 = {ap}^2 + {cp}^2\implies({(-2)}^2+6^2+4^2+2^2) = ({(x-2)}^2 +{(y-5)}^2) + (2^2+{(8-x)}^2+{(9-y)}^2+{(-2)}^2 )$
$60 = 2x^2+2y^2-20x-28y+182\implies x^2+y^2-10x-14y+61=0$
$2. \ \ \triangle bcp:\ \ \ \ \ {bc}^2 = {bp}^2 + {cp}^2\implies(2^2+4^2+2^2+2^2) = ({(x-2)}^2 +{(y-5)}^2) + (2^2+{(8-x)}^2+{(9-y)}^2+{(-2)}^2 )$
$28 = 2x^2+2y^2-20x-28y+190\implies x^2+y^2-12x-16y+95=0$
From $1,2$, get: $-2x -2y +34 = 0\implies x +y -17=0$.
But, how to proceed it further to find coordinates of $p$ is unclear.
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Having $p=(3,x,y,1)$ does not reflect that $p$ is on the line $ab$. However, parametric equation of an arbitrary line passing through arbitrary points $a,\,b$ in Euclidean space is $x=a+t(b-a)$ with some parameter $t\in \mathbb{R}$, hence
$$p=a+t(b-a)=(3,2,5,1)+t\,(0,2,2,0)$$
or, more compactly $$p=(3,2+2t,5+2t,1)$$ and your further computations will succeed as we have only one equation
$$(b-a)\cdot(p-c)=0\;(\Leftrightarrow (b-a)\perp (p-c))$$
in other words, the two equations in the OP solution are equivalent and do not give solutions to two variables.
$$(0,2,2,0)\cdot((3,2+2t,5+2t,1)-(5,8,9,3))=0$$
$$(0,2,2,0)\cdot(-2,2t-6,2t-4,-2)=0$$
$$2\cdot(2t-6)+2\cdot(2t-4)=0$$
$$t=\frac{5}{2}$$
$$p=(3,7,10,1)$$
Verification, for a case of a typo: $(b-a)\cdot(p-c)=0$.
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|
Problem with Summation of series Question: What is the value of $$\frac{1}{3^2+1}+\frac{1}{4^2+2}+\frac{1}{5^2+3} ...$$ up to infinite terms?
Answer: $\frac{13}{36}$
My Approach:
I first find out the general term ($T_n$)$${T_n}=\frac{1}{(n+2)^2+n}=\frac{1}{n^2+5n+4}=\frac{1}{(n+4)(n+1)}=\frac{1}{3}\left(\frac{1}{n+1}-\frac{1}{n+4}\right)$$
Using this, I get,$$T_1=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}\right)$$ $$T_2=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}\right)$$ $$T_3=\frac{1}{3}\left(\frac{1}{4}-\frac{1}{7}\right)$$
I notice right away that the series does not condense into a telescopic series. How do I proceed further?
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You have solved it already. Recognise, that you can factor out $\frac 13$ from all terms. You have an infinite terms of +/- fractions, from which the first positive three ($\frac 12$, $\frac 13$, $\frac 14$) remains in the sum, every other is cancelled out by a negative counterpart with a 3 step gap. Therefore the end result is $\frac 13 (\frac 12 + \frac 13 + \frac 14)$.
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Solve the equation $\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$ Problem
Solve the equation $$\frac{1}{x^2+11x-8} + \frac{1}{x^2+2x-8} + \frac{1}{x^2-13x-8} = 0$$
What I've tried
First I tried factoring the denominators but only the second one can be factored as $(x+4)(x-2)$.
Then I tried substituting $y = x^2 - 8$ but that didn't lead me anywhere.
Where I'm stuck
I don't know how to start this problem. Any hints?
P.S. I would really appreciate it if you give me hints or at least hide the solution. Thanks for all your help in advance!
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You started very well. To make things easier, set $A=x^2+7x-8$ (*) and the equation rewrites
$$\frac{1}{A+4x} + \frac{1}{A-5x} + \frac{1}{A-20x} = 0.$$
Denominators are not allowed to be zeros. We solve
$$(A-5x)(A-20x)+(A+4x)(A-20x)+(A+4x)(A-5x)=0$$
or equivalently
$$3A^2-42Ax=0,$$ or even
$$3(x^2+7x-8)(x^2-7x-8)=0,$$
which is easy to finish.
(*) I noticed that $x=1$ satisfies, and decided to make profit from it.
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Given the equation $\alpha \mathbf{v} + \mathbf{v}\times\mathbf{a} = \mathbf{b}$, solve for $\mathbf{v}$. I'm reading a textbook at the moment that provides the following linear equation,
$$
\alpha \mathbf{v} + \mathbf{v}\times\mathbf{a} = \mathbf{b},
$$
and asks to solve for $\mathbf{v}$. The form of $\mathbf{v}$ is given as
$$
\mathbf{v} = \frac{\alpha^2 \mathbf{b} - \alpha (\mathbf{b} \times \mathbf{a}) + (\mathbf{a}\cdot\mathbf{b})\mathbf{a}}{\alpha(\alpha^2+\lvert \mathbf{a} \rvert^2)}.
$$
It's easy enough to verify that this is the correct solution. However, I can't figure out how I'd solve for $\mathbf{v}$ if given just the original equation.
Are there any general approaches to solving this kind of equation systematically?
Edit: $\mathbf{a}, \mathbf{b}$ and $\mathbf{v}$ are all vectors, whereas $\alpha$ is a scalar such that $\alpha \neq 0$.
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Taking cross product with $\mathbf{a}$ on both sides, we get,
\begin{align*}
&\alpha \mathbf{v} + \mathbf{v}\times \mathbf{a} = \mathbf{b}\\
\implies &\alpha(\mathbf{v}\times \mathbf{a})+(\mathbf{v}\times \mathbf{a})\times \mathbf{a}=\mathbf{b}\times \mathbf{a}\\
\implies &\alpha(\mathbf{b}-\alpha \mathbf{v})+(\mathbf{v}\cdot \mathbf{a})\mathbf{a}-|a|^2\mathbf{v}=\mathbf{b}\times \mathbf{a}\\
\implies &\alpha \mathbf{b}-\alpha^2\mathbf{v}+\dfrac1\alpha (\mathbf{b}\cdot \mathbf{a})\mathbf{a}-|a|^2\mathbf{v}=\mathbf{b}\times \mathbf{a}&&\Big(\text{Using }\alpha (\mathbf{v}\cdot \mathbf{a})=\mathbf{b}\cdot \mathbf{a}\Big)
\end{align*}
Now solve for $\mathbf{v}$ directly.
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|
Integrate $\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln{x}} \mathop{dx}$ Insane integral $$\int_0^1 \frac{x^2-1}{\left(x^4+x^3+x^2+x+1\right)\ln x} \mathop{dx}$$
I know $x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}$ but does it help? I think $u=\ln{x}$ might be necessary some point.
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Write this as:
$$\int_{0}^1\frac{(x^2-1)(x-1)\,dx}{(x^5-1)\ln x}$$
Note that $$\int_0^1 x^y\,dy = (x-1)/\ln x$$
Replacing this and converting the problem into a double integral gives
$$\int_{0}^1\int_0^1\frac{(x^2-1)}{(x^5-1)}x^y\,dx\, dy$$
Writing $\frac{1}{1-x^5} = \sum_{k=0}^\infty x^{5k}$
This becomes
$$\int_{0}^1\int_0^1(1-x^2)x^y\sum_{k=0}^\infty x^{5k}\, dx\, dy$$
Interchanging the summation and integration
$$ = \sum_{k=0}^\infty \int_{0}^1\int_0^1(1-x^2)x^yx^{5k}\,dx\,dy$$
Integrating with respect to $x$,
$$ = \sum_{k=0}^\infty \int_{0}^1 \left(\frac{1}{5k+y+1} - \frac{1}{5k+y+3}\right)\,dy$$
Integrating this, we get
$$I = \sum_{k=0}^\infty \ln \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)}$$
Using the identity provided by @Nanayajitzuki in the comments:
$$\ln \prod_{k=0}^\infty \frac{(5k+2)(5k+3)}{(5k+1)(5k+4)} = \ln\left(\frac{\Gamma(1/5)\Gamma(4/5)}{\Gamma(2/5)\Gamma(3/5)}\right)$$
Using Euler's reflection formula, $\Gamma(z)\Gamma(1-z) = \pi /\sin (\pi z)$
$$ = \ln \frac{\sin (2\pi/5)}{\sin (\pi/5)}$$
$$ = \ln (2\cos (\pi/5))$$
$$ = \ln (\phi)$$
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|
Find value of $\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$ My attempt :
$$\dfrac{\left(1+\tan^2\dfrac{5\pi}{12}\right)\left(1-\tan^2\dfrac{\pi}{12}\right)}{\tan\dfrac{\pi}{12}\tan\dfrac{5\pi}{12}}$$
Change into variable form
$$\dfrac{(1+a^2)(1-b^2)}{ab}$$
$$\dfrac{1+a^2-b^2-a^2b^2}{ab}$$
I'm stuck here also I don't think this is the correct way.
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$$\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{11\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{17\pi}{12}}$$
$$=\dfrac{(1+\tan^2\frac{5\pi}{12})({1-\tan^2\frac{\pi}{12}})}{\tan\frac{\pi}{12}\tan\frac{5\pi}{12}}$$
$$=\dfrac{4}{\left(\dfrac{2\tan\frac{5\pi}{12}}{(1+\tan^2\frac{5\pi}{12})}\right)\left(\dfrac{2\tan\frac{\pi}{12}}{{1-\tan^2\frac{\pi}{12}}}\right)}$$
Use trig. identity, $\frac{2\tan\theta}{1+\tan^2\theta}=\sin2\theta$, $\frac{2\tan\theta}{1-\tan^2\theta}=\tan2\theta$,
$$=\dfrac{4}{\left(\sin\left(2\frac{5\pi}{12}\right)\right)\left(\tan\left(2\frac{\pi}{12}\right)\right)}$$
$$=\dfrac{4}{\left(\sin\frac{5\pi}{6}\right)\left(\tan\frac{\pi}{6}\right)}$$
$$=\dfrac{4}{\left(\frac12\right)\left(\frac{1}{\sqrt 3}\right)}$$
$$=8\sqrt3$$
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|
Solve the inequality $|3x-5| - |2x+3| >0$. In order to solve the inequality $|3x-5| - |2x+3| >0$, I added $|2x+3|$ to both sides of the given inequality to get $$|3x-5| > |2x+3|$$ Then assuming that both $3x-5$ and $2x+3$ are positive for certain values of $x$, $$3x-5 > 2x+3$$ implies $$x>8$$ If $3x-5$ is positive and $2x-3$ is negative for certain values of $x$, then $$3x-5 > -2x-3$$ implies $$5x >2$$ implies $$x > \dfrac{2}{5}$$ I'm supposed to get that $x < \dfrac{2}{5}$ according to the solutions, but I'm not sure how to get that solution.
|
$\left|3x-5\right|$ is negative for $x<-\frac{3}{2}$ and $-\frac{3}{2}\le x<\frac{5}{3}$. Positive for $x\ge \frac{5}{3}$.
For $\left|2x+3\right|$ you have in the same intervals the signs: $-$, $+$ and $+$.
Hence for $x<-\frac{3}{2}$ you have $5-3x-\:2x-3\:>\:0 \iff x<\frac{2}{5}$. and so on and so removing the absolute values in relation to the intervals where they are positive or negative.
Combining the intervals you will have:
$$x<-\frac{3}{2}\quad \mathrm{or}\quad \:-\frac{3}{2}\le \:x<\frac{2}{5}\quad \mathrm{or}\quad \:x>8$$ and merge overlapping the intervals
$$x<\frac{2}{5}\quad \mathrm{or}\quad \:x>8$$
|
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|
Help with the last step in solving $\lim_{x\to0}\frac{(1+\sin x +\sin^2 x)^{1/x}-(1+\sin x)^{1/x}}x$ I managed to solve part of this limit but can't get the final step right. Here's the limit:
$$
\lim_{x\to0} {
\frac
{
\left(
1+\sin{x}+\sin^2{x}
\right)
^{1/x}
-
\left(
1+\sin{x}
\right)
^{1/x}
}
{
x
}
}
$$
Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
x
}
\right]
}
{
x
}
}
$$
Multiplying and dividing by the right terms:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}+\sin^2{x}
}
{
x
}
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Grouping the $\sin{x}$ in the first exponential function:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
$$
Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$
We should end up with:
$$
\lim_{x\to0} {
\frac
{
\exp
\left[
\frac
{
\log{\left(1+\sin{x}+\sin^2{x}\right)}
}
{
\sin{x}+\sin^2{x}
}
\frac
{
\sin{x}
}
{
x
}
\left(1+\sin{x}\right)
\right]
-
\exp
\left[
\frac
{
\log{\left(1+\sin{x}\right)}
}
{
\sin{x}
}
\frac
{
\sin{x}
}
{
x
}
\right]
}
{
x
}
}
=
\frac
{
\exp{\left[1(1)(1)\right]}
-
\exp{\left[1(1)\right]}
}
{
0
}
= \frac{e-e}{0}
= \frac{0}{0}
$$
Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.
|
Your second last equality is not justified: You cannot take limit of enumerator and denominator separately as the limit of the denominator is (trivially) equal to 0. Try the appropriate version of l'Hospitals rules (which helps, because enumerator and denominator (as you have formally shown) converge to $0$.
|
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|
Finding real $(x,y)$ solutions that satisfies a system of equation. I was given:
$x + y^2 = y^3 ...(i) \\ y + x^2 = x^3...(ii)$
And was asked to find real $(x,y)$ solutions that satisfy the equation.
I substracted $(i)$ by $(ii)$:
$x^3 - y^3 + y^2 - x^2 + x - y = 0$
Then factored it out so I have:
$(x-y)(x^2 + xy + y^2 - x - y + 1) = 0$
Multiplying it by two, I get:
$(x-y)(2x^2 + 2xy + 2y^2 - 2x - 2y + 2) = 0 \\
(x-y)((x^2 - 2x + 1) + (y^2 - 2y + 1) + (x^2 + 2xy + y^2)) = 0 \\
(x-y)((x-1)^2 + (y-1)^2 + (x+y)^2) = 0$
I noticed that a solution exists only if $x=y$ because there are no real solutions for $x$ and $y$ that satisfies $(x-1)^2 + (y-1)^2 + (x+y)^2 = 0$.
Substituting $x=y$ into the first equation, I get:
$y(y^2-y-1)=0$ where the roots are $y= 0, \frac{1+\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2}$. Hence, the real solutions of $(x,y)$ that satisfy are:
$(x,y) = (0,0), (\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2}), (\frac{1-\sqrt{5}}{2}, \frac{1-\sqrt{5}}{2})$.
What I would like to ask is: Is there a better way to solve the question? It's from a local university entrance test, where this kind of questions are aimed to be done in < 3 minutes. It took me a while to manipulate the algebraic stuffs above.
Someone in a local forum said something about symmetric systems which says that the solution does not exist for $x \neq y$. How do I know if the equation is a symmetric one? (Never heard of something before throughout high school here...) I would love to see a resource for this!
|
You can begin by noting that the $2$ functions are inverses of each other (and only involve odd non-zero exponents). Using the fact that inverse functions are reflections in the line $y=x$, we can now see that the intersection points must be along the line $y=x$. Substituting $y$ into $x$ or vice-versa, we obtain the equation you get and obtain the solutions you got. That would only take about 3 minutes. I hope that helps :)
|
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|
Proving $\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$ algebraically
The question is to prove that for any positive real numbers $x$, $y$ and $z$,
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
So I decided to do some squaring on both sides and expanding:
$$\sqrt{x^2-xz+z^2} + \sqrt{y^2-yz+z^2} \geq \sqrt{x^2+xy+y^2}$$
$$x^2 -xz+z^2 +y^2-yz+z^2+2\sqrt{(x^2-xz+z^2)(y^2-yz+z^2)} - x^2 - xy - y^2 \geq 0$$
$$2\sqrt{(x^2-xz + z^2)(y^2-yz+z^2)}\geq xz+yz+xy-2z^2$$
Squaring both sides yields
$$4(x^2-xz+z^2)(y^2-yz+z^2)\geq(xy+xz+yz-2z^2)^2$$
After some expanding it becomes
$$(xy-xz-yz)^2 \geq 0$$
which completes the proof.
However, I want to ask whether the squaring in the $3^{rd}$ - $4^{th}$step is problematic. Since it is possible for $xy + xz+ yz -2z^2$ to be negative. Squaring both sides will invert the sign. Do I have to analyse the positive and negative case here seperately?
|
A Geometric Proof:
Take a point $O$ in the plane and consider three line segments (this can be done since $x,y,z$ are positive real numbers) $OA, OB, OC$ with $$|OA|=x,|OB|=z,|OC|=y$$ and $$\angle AOB=60^{\circ},\angle BOC=60^{\circ}$$ therefore $\angle AOC=120^{\circ}$. In the triangle $\Delta ABC$ we have $$AB+BC\geq AC\tag{1}$$ Again by cosine rule $$AB=\sqrt{x^2-xz+z^2}\\BC=\sqrt{y^2-yz+z^2}\\AC=\sqrt{x^2+xy+y^2}$$ Hence we get the desired inequality.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
Prove that $\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{n^2} > \frac{3n}{2n+1}$ for all $n \geq 2$ by induction
I tried to add $\frac{1}{(k+1)^2}$ to both sides of this inequality assuming it's true for $n = k$, but eventually it resulted in a complex expression in the second side:
$$
\frac{3k}{2k+1} + \frac{1}{(k+1)^2} = \frac{3k^3+6k^2+5k+1}{(k+1)^2 (2k+1)}
$$
that I'm not sure how to finish the proof using it.
Am I going the right way? If so, how can I finish this proof?
and is there an easier way to prove this by induction without facing such confusing expressions?
Thanks.
|
When $n=2$, it directly works.
Assume it works for $n=k\ge 2$, and try to show that for $n=k+1$.
$1+...+\frac{1}{k^2}>\frac{3k}{2k+1}$ is given, so it suffices to show $\frac{1}{(k+1)^2}\ge\frac{3(k+1)}{2(k+1)+1}-\frac{3k}{2k+1}=\frac{3}{4(k+1)^2-1}\Leftrightarrow (k+1)^2\ge 1$, which is obvious.
|
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|
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$ I was reading a solution in which author used this identity without giving any hint that how it comes ...{maybe it is obvious}
$2^{3^{n}}+1=(2+1)\left(2^{2}-2+1\right)\left(2^{2 \cdot 3}-2^{3}+1\right) \cdots\left(2^{2 \cdot 3^{n-1}}-2^{3^{n-1}}+1\right)$
by using expansion of $a^n+b^n$ i get
$2^{3^{n}}+1=(2+1)(2^{3^n-1}-2^{3^n-2}+....+1)$
but how we get above ???
|
For those of you who would like to see the full version of this:
Applying the factorization $x^3+1=(x+1)(x^2-x+1)$ to $x=a^{3^k}$ gives
$$ a^{2 \cdot 3^k}-a^{3^k}+1 = \frac{a^{3^{k+1}}+1}{a^{3^k}+1}. $$
Therefore,
$$ \prod_{k=0}^{n-1} \left(a^{2 \cdot 3^k}-a^{3^k}+1\right) = \prod_{k=0}^{n-1} \frac{a^{3^{k+1}}+1}{a^{3^k}+1} = \frac{a^{3^n}+1}{a^{3^0}+1} = \frac{a^{3^n}+1}{a+1}. $$
The second equality in the displayed equation above is an example of telescoping product - each denominator cancels the following numerator, leaving us the first numerator and the last denominator.
|
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|
Find $\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$ Original question is $$\sum_{n=1}^{\infty} \frac{1}{\prod_{i=0}^{k} \left(n+i\right)}$$
I got it down to $$\sum_{n=1}^{\infty} \frac{(n-1)!}{(k+n)!}$$
Here I am confused. Possible fraction decomposition but its ugly! Maybe this approach is not good? Ideas?
Answer is $$\frac{1}{k \cdot k!}$$
I want to know how to proceed with my work though
|
\begin{align*}
\sum_{n=1}^\infty \frac{1}{n(n+1)...(n+k)} &= \frac{1}{k} \sum_{n=1}^\infty \frac{k}{n(n+1)...(n+k)} \\
&= \frac{1}{k} \sum_{n=1}^\infty \left[ \frac{1}{n(n+1)...(n+k-1)} - \frac{1}{(n+1)...(n+k)} \right], \\
\end{align*}
and this series telescopes so that every minus cancels with a plus, and we are left with only the first plus term, when $n = 1$:
$$\frac{1}{k} \frac{1}{1(1+1)...(1+k-1)} = \frac{1}{k * k!}$$
|
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|
Closed form of $\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}}$ How to prove the following formula,
$$\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{x}}}} =
\begin{cases}
2 \cos\Big[\frac{1}{2^{n}}\Big(\pi + \arctan{\frac{\sqrt{x(4-x)}}{x-2}}\Big)\Big] & 0 < x < 2 \\
2 \cos\Big[\frac{1}{2^{n}}\arctan{\frac{\sqrt{x(4-x)}}{x-2}}\Big] & 2 < x \leq 4 \\
2 \cosh\Big[\frac{1}{2^{n}}\text{arctanh}{\frac{\sqrt{x(x-4)}}{x-2}}\Big] & x \geq 4
\end{cases}
\
$$
where $x \in \mathcal{R}$ and $n$ is the number of time the square root symbol $\sqrt{\text{ }\text{ }}$ appears.
This formula is a generalization of the following formula, which can be found here and here, and which can be obtained by taking the above general formula in the limit $x = 2$.
$$\sqrt{2 + \sqrt{2 + ... + \sqrt{2 + \sqrt{2}}}} = 2\cos\Big(\frac{\pi}{2^{n+1}}\Big)$$
|
Hint If $a=2\cos(y)$ then
$$2+a= 2 + 2\cos(y)= 4 \cos^2(\frac{y}{2}) \,.$$
So write $\sqrt{x}=2 \cos(\alpha)$ and use the above formula:
$$2 +\sqrt{x}= (2 \cos(\frac{\alpha}{2}))^2 \\
2+\sqrt{2+\sqrt{x}}=2+2\cos(\alpha/2)= (2\cos(\alpha/4))^2 \\
...$$
|
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|
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
How many unordered pairs of positive integers $(a,b)$ are there such that $\operatorname{lcm}(a,b) = 126000$?
Attempt:
Let $h= \gcd(A,B)$ so $A=hr$ and $B=hp$, and $$phr=\operatorname{lcm}(A,B)=3^2\cdot 7\cdot 5^3 \cdot 2^4\,.$$ Let $p = 3^a5^b7^c2^d$ and $r = 3^e 5^f 7^g 2^s$. Notice, that given $p$ and $r$, $h$ is determined, so we can count $p$ and $r$. Multiplying $p$ and $r$ we get $$pr = 3^{(a+e)} 5^{(b+f)} 7^{(c+g)} 2^{(d+s)}\,,$$ and so $a+e = 0,1,2$.
For the first case we have $0+1 = 1$ possibility, similarly $2$ and $3$ for the other cases, so the total number is $6$. For $b+f$ we have $b + f = 0,1,2,3$ giving $10$ options. Similarly for $c + g$ we have $3$ choices and for $d + h$ we have $$1+2+3+4+5 = 15$$ choices. Multiplying these together, we get $$15\cdot 3\cdot 6\cdot 10 = 60\cdot 45 = 2700\,,$$ which is not equal to the given answer of $473$.
Edit: sorry for the weird variables. I think I've fixed everything, if not, please do point it out
|
Think it through: to find the lowest common divisor of $a$ and $b$ you take the prime factorization of $a$ and $b$. And find the number whose prime factorization consists of the prime factors of $a$ and of $b$ and raising them to the higher power they are raised to in $a$ and $b$. In other words:
If $\{p_i\}$ are the prime factors of $a$ and/or $b$ and $a = \prod p_i^{k_i}$ ($k_i$ may be equal to $0$ if $p_i|b$ but $p_i\not \mid a$) and $b = \prod p_i^{j_i}$ (similarly $j_i$ may be $0$ if $p_i\not \mid b$ but $p_i|a$) then $\operatorname{lcm}(a,b) = \prod p_i^{\max (k_i,j_i)}$.
so if $\operatorname{lcm}(a,b)=126000 = 2^4*3^2*5^3*7$ then $a= 2^{k_1}*3^{k_2}5^{k_3}7^{k_4}$ and $b=2^{j_1}*3^{j_2}5^{j_3}7^{j_4}$ where $\max(k_i,j_i) = 4,2,3,1$
So it becomes a combinatorial matter of counting these.
This is easier if we consider ordered pairs.
One of $a$ or $b$ must have $2^4$ divide it. There is $1$ way that both $a$ and $b$ have $2^4$ divide it. There are $2$ ways that $2^4$ divides $a$ or $b$ but not the other. And the $2^0,2^1,2^2,2^3$ are four ways that $2$ may divide then others. So there are $2*4 + 1=9$ we can factor up $2^4$ between $a$ and $b$.
We can do the same for the factors $3^2,5^3, 7^1$ to get $2*2+1=5;2*3+1=7;2*1+1=3$ ways, respectively to distribute those factors.
So there are $9*5*7*3 = 945$ *ordered pairs $(a,b)$ where $\operatorname{lcm}(a,b) =126000$.
But we need unordered pairs. So we take the number of pairs where $(a,b)\ne (b,a)$ (ie; $a\ne b$) and divide by $2$. And we take the pairs where $(a,b)=(b,a)$ (ie; $a=b$) and add those.
But if $a=b$ and $\operatorname{lcm}(a,b)=\operatorname{lcm}(a,a) = 126000$ then $a=b=126000$ and there is only one such pair.
So there are $945-1 = 944$ pairs with $a\ne b$. So that is $\frac {944}2=472$ unordered pairs where $a\ne b$ and one pair where $a=b= 126000$. So there are $473$ pairs of unordered $(a,b)$ where $\operatorname{lcm}(a,b)=126000$.
|
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|
How to solve $\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx} $? $$\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, \mathrm{dx}$$
I have tried $u$-substitution and multiplying by the conjugate and then apply $u$-substitution. For the $u$-substitution, I have set $u$ equal to each square root term, set $u$ equal to the entire denominator, and set $u$ equal to each expression in the radical.
However, all my attempts have just made the integral more complex without an obvious way to simplify. Can someone provide insight please? Thank you.
|
$$I=\int\frac{1}{\sqrt {2x} - \sqrt {x+4}} \, dx$$
Get rid of the denominator
$$u=\sqrt {2x} - \sqrt {x+4}\implies x=3 u^2+2 \sqrt{2}u \sqrt{u^2+4 }+4\implies \frac{dx}{du}=\frac{\sqrt{2} \left(4 u^2+8 \right)}{\sqrt{u^2+4 }}+6 u$$
$$I=\int6 du+\int\frac{4 \sqrt{2} u}{\sqrt{u^2+4}}du+\int\frac{8 \sqrt{2}}{u\sqrt{u^2+4} }du$$
$$I=6u+4 \sqrt{2} \sqrt{u^2+4}-4 \sqrt{2} \log \left(\frac{\sqrt{u^2+4}+2}{u}\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to find indefinte integral $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using substitution of $u=\cos(\frac{x^2}{4})$?
Can someone help me find $\int_{0}^{\sqrt\pi} 2x\cos(\frac{x^2}{4})dx$ using the substitution $u=\cos(\frac{x^2}{4})$ step-by-step?
All I've done so far is:
$$\frac{du}{dx}=-\frac{1}{2}x\sin(\frac{x^2}{4})$$
and
$$x=\sqrt\pi \longrightarrow u=\frac{\sqrt2}{2}$$
$$x=0 \longrightarrow u=1$$
$$\therefore \int_{1}^{\frac{\sqrt2}{2}}\frac{-4u}{\sqrt{1-u^2}}du$$
This is where I don't know how to continue, can ayone help me understand how to finish this off?
|
From your steps
\begin{align*}
&\int_{1}^{\frac{1}{\sqrt{2}}} \frac{-4u}{\sqrt{1 - u^2}}du\\
=&~2 \int_{0}^{\frac{1}{2}} \frac{dt}{\sqrt{t}} \quad \text{on the substituion } (1 - u^2) = t\\
=&~2 \int_{0}^{\frac{1}{\sqrt{2}}}\frac{w}{w}d\omega\quad \text{on the substituition } t = \omega^2\\
=&~4 \bigg[\frac{1}{\sqrt{2}}\bigg] = 2\sqrt{2}
\end{align*}
|
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|
Computing $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} + 2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$ How can I compute the sum $2 \binom{n}{0} + 2^2 \frac{\binom{n}{1}}{2} +
2^3 \frac{\binom{n}{2}}{3} + \cdots + 2^{n+1} \frac{\binom{n}{n}}{n+1}$? I think I should expand $(1+ \sqrt{2})^n$ or something like this and then find some kind of linear recurrence, but I'm not sure.
|
Your expression is $\displaystyle\sum_{k=0}^n \frac{x^{k+1}}{k+1} \binom{n}{k}$ evaluated at $x=2$.
Hint. Can you think of where else $\displaystyle\frac{x^{k+1}}{k+1}$ shows up (specifically, in calculus)?
|
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"url": "https://math.stackexchange.com/questions/3764054",
"timestamp": "2023-03-29T00:00:00",
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|
Find the recursive formula for $I_m:=\int \frac{dx}{\sqrt{x^2+1}^m},\enspace \,m\in\mathbb{N}$ Find the recursive formula for:
$$I_m:=\int \frac{dx}{\sqrt{x^2+1}^m}\,\,,\,\,\,\,m\in\mathbb{N}$$
My attempt:
$$I_m=\int\frac{dx}{\sqrt{x^2+1}^m}=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\int \frac{x^2}{\sqrt{x^2+1}^m(x^2+1)}dx$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\int \left(1-\frac{1}{x^2+1}\right)\frac{1}{\sqrt{x^2+1}^m} dx $$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\left(\int\frac{1}{\sqrt{x^2+1}^m}dx- \int\frac{1}{\sqrt{x^2+1}^m(x^2+1)} dx \right)$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+m\left(\int\frac{1}{\sqrt{x^2+1}^m}dx- \int\frac{1}{\sqrt{x^2+1}^{m+2}} dx \right)$$
$$=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+mI_m-mI_{m+2}$$
$$\Longleftrightarrow mI_{m+2}=\left[\frac{x}{\sqrt{x^2+1}^m}\right]+(m-1)I_m$$
So we finally get:
$$ I_{m+2}=\frac{1}{m}\left[\frac{x}{\sqrt{x^2+1}^m}\right]+\frac{m-1}{m}I_m$$
Is this correct? Would be great if someone could look over it :) Thank you a lot as always
|
Another way:
$$\dfrac{d\dfrac x{(x^2+1)^{n/2}}}{dx}=\dfrac1{(x^2+1)^{n/2}}-\dfrac{nx}{2(x^2+1)^{n/2+1}}\cdot2x$$
$$=\dfrac1{(x^2+1)^{n/2}}-n\cdot\dfrac{1+x^2-1}{(x^2+1)^{n/2+1}}$$
Integrate both sides w.r.t $x,$
$$\implies nI_{n+2}-(n-1)I_n=\dfrac x{(x^2+1)^{n/2}}+K$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
What should $n$ be equal to, so that $5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$ is completely divisible by $19$? What should $n$ be equal to, so that the number:
$$5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}$$
is completely divisible by 19? I broke it into this:
$$20\cdot 2^{n}\cdot 25^{n}+18\cdot 3^{n}\cdot 4^{n}$$ But what should i do next?
|
For any natural $n$ we obtain:
$$\begin{aligned}5^{2n+1}2^{n+2} + 3^{n+2}2^{2n+1}&=2^{n+1}\left(10\cdot5^{2n}+9\cdot6^n\right) \\
&= 2^{n+1}\left(10\cdot(6+19)^{n}+9\cdot6^n\right) \\
& \equiv2^{n+1}\left(10\cdot6^n+9\cdot6^n\right) \pmod{19} \\ &\equiv 0 \pmod{19}.\end{aligned}$$
|
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|
Contradiction on an Inequality The problem on which I have a problem is this-
Let a, b, c be non-negative real numbers. Prove that $$ \sum_{cyc} {a^2-bc \over 2a^2+b^2+c^2} \ge 0 $$
While solving and after some resolution, we get $$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le 3 $$
And by C-S, we have,
$$ \sum_{cyc} {(a+b)^2 \over a^2+b^2+2c^2} \le \sum_{cyc} {2(a^2+b^2) \over a^2+b^2+2c^2} $$
It rests to prove that
$$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} \le {3 \over 2} $$
By cyclic substitutions of $x$ for $b^2+c^2$, we get,
$$ \sum_{cyc} {a^2+b^2 \over a^2+b^2+2c^2} = \sum_{cyc} {(a^2+b^2) \over (c^2+a^2)+(b^2+c^2)} = \sum_{cyc} {z \over x+y} \le {3 \over 2} $$
But by the Nesbitt's Inequality,
$$ \sum_{cyc} {z \over x+y} \ge {3 \over 2} !$$
Can anybody explain me where's the mistake and the correction?
Thanks!
|
Another solution.
By your work we need to prove that:
$$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq3,$$ which is true by C-S:
$$\sum_{cyc}\frac{(a+b)^2}{a^2+b^2+2c^2}\leq\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{b^2}{b^2+c^2}\right)=\sum_{cyc}\left(\frac{a^2}{a^2+c^2}+\frac{c^2}{c^2+a^2}\right)=3.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A trick system of equations \begin{cases}
x+y+\dfrac{x^2}{y^2}=7\\
\dfrac{(x-y)x^2}{y^2}=12
\end{cases}
I don't have any idea to solve this. I tried to subtract, add and multiply the given equations, but nothing help me to find $x$ or $y$.
Can someone help me to solve this system?
Thanks for attention.
|
The second equation tells you that $$\frac{x^2}{y^2}=\frac{12}{x-y}$$
(The possibility $x-y=0$ is ruled out, because this would violate the second equation.)
If we substitute this in the first equation, we get
$$x+y+\frac{12}{x-y}=7$$
which implies
$$(x+y)(x-y)+12=7(x-y)$$
$$12=7(x-y)-(x-y)(x+y)$$
$$12=(x-y)[7-(x+y)]$$
I was trying to find integer solutions. $12 = 6.2 \textrm{ or } 2.6 \textrm{ or } 12.1 \textrm{ or } 1.12 \textrm{ or } 4.3 \textrm{ or } 3.4$ (6 different cases).
The first case results in $x=5.5$ and $y=-0.5$. (This is very easy)
The second case results in $x=1.5$ and $y=-0.5$.
The third case results in $x=9$ and $y=-3$.
The fourth case results in $x=-2$ and $y=-3$.
The fifth case results in $x=4$ and $y=0$ (but $y \neq 0)$.
The sixth case results in $x=3$ and $y=0$ (but $y \neq 0)$.
We can do the same with the negative integer factors of 12. That would give us extra solutions.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
|
A variation of other answers (that more closely parallels a common pattern when the numerator and denominator are polynomials) is "big part factoring". \begin{align*}
\lim_{x \rightarrow 0} \frac{12^x - 4^x}{9^x-3^x}
&= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{4}{12} \right) ^x \right)}{9^x \left( 1-\left( \frac{3}{9} \right) ^x \right) } \\
&= \lim_{x \rightarrow 0} \frac{12^x \left(1 - \left( \frac{1}{3} \right) ^x \right)}{9^x \left( 1-\left( \frac{1}{3} \right) ^x \right) } \\
&= \lim_{x \rightarrow 0} \frac{12^x }{9^x} \\
&= \frac{1}{1} \\
&= 1 \text{.}
\end{align*}
|
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"timestamp": "2023-03-29T00:00:00",
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|
Positive integer solutions to $\frac{1}{a} + \frac{1}{b} = \frac{c}{d}$ I was looking at the equation $$\frac{1}{a}+\frac{1}{b} = \frac{c}{d}\,,$$ where $c$ and $d$ are positive integers such that $\gcd(c,d) = 1$.
I was trying to find positive integer solutions to this equation for $a, b$, given any $c$ and $d$ that satisfy the above conditions. I was also trying to find whether there are additional requirements on $c$ and $d$ so that positive integer solutions for $a$ and $b$ can even exist.
I found that this equation simplifies to $abc - ad - bd = 0$ so that $abc = d(a+b)$.
Also, since the equation is equivalent to $a+b = ab(\frac{c}{d})$, this means $a$ and $b$ are the roots of the quadratic $dx^2-abcx+abd = 0$ since their product is $ab$ and their sum is $a+b = ab(\frac{c}{d})$.
However, after I analyzed the quadratic I just ended up with $a = a$ and $b = b$.
Any ideas on how to solve this further?
Again, I need to find all the conditions on the positive integers $c$ and $d$ (where $\gcd(c,d) = 1$) such that positive integer solutions for $a, b$ can exist. And then also find the positive integer solutions for $a$ and $b$ given that those conditions are satisfied.
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Try some examples. I suggest researching Simon's Favorite Factoring trick, it helps with these types of questions. Asking on artofproblemsolving.com is a great idea, Simon is a real person who invented this factoring trick, and it's on that website.
I'll give you one example. Let's take $\frac{1}{a}+\frac{1}{b}=\frac{1}{4}$. Note that $\frac{a+b}{ab}=\frac{1}{4}$. So, $4a+4b=ab$, meaning that $ab-4a-4b=0$, so $(a-4)(b-4)=16$. To solve this in the integers, just find the factor pairs of $16$ and solve for $a,b$
Hope this helped you!
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is value of this integral? $\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^{2})}dx$
What is value of this integral $$I=\int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)}dx$$
My work :
\begin{align*}I&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)(10+10x^2)}{(9+x^2)(1+9x^2)}dx\\
&=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{1+9x^2}dx+\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx\\
&=j_{1}+j_{2}+j_{3}\\
\end{align*}
$$j_{1}=\int_{0}^{\infty}\frac{\log(1+4x^2)}{9+x^2}dx=\sum_{n=0}^{\infty}\frac{(-1)^n(2^{2(n+1)})}{n+1}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
$$j_{2}=\log(4)\int_{0}^{\infty}\frac{1}{1+(3x)^2}dx+\sum_{n=0}^{\infty}\frac{(-1)^{n}}{(n+1)(2^{2(n+1)})}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx=\frac{\log(4)\pi}{6}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{1+9x^2}dx$$
$$j_{3}=\int_{0}^{\infty}\frac{\log(4+x^2)}{9+x^2}dx=\frac{\log(4)\pi}{54}+\sum_{n=0}^{\infty}\frac{(-1)^n}{(n+1)2^{2(n+1)}}\int_{0}^{\infty}\frac{x^{2(n+1)}}{9+x^2}dx$$
Wait for a review to find solutions to this
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Consider using differentiation under the integral sign. Parameterize the integral as the following:
$$I(a)=\int_0^{\infty} \frac{\ln{\left(1+ax^2\right)}}{9+x^2} + \frac{\ln{\left(a+x^2\right)}}{1+9x^2}+ \frac{\ln{\left(a+x^2\right)}}{9+x^2} \; \mathrm{d}x$$
The integral in question is $I(4)$. First, differentiate $I(a)$ with respect to $a$: \begin{align*}
I'(a)&=\int_0^{\infty} \frac{x^2}{(9+x^2)(1+ax^2)}+\frac{1}{(1+9x^2)(a+x^2)}+\frac{1}{(9+x^2)(a+x^2)} \; \mathrm{d}x \\
&=\int_0^{\infty} -\frac{10 (a - 1)}{(a - 9) (9 a - 1) (a + x^2)} + \frac{2 (9 a - 41)}{(a - 9) (9 a - 1) (x^2 + 9)} + \frac{9}{(9 a - 1) (9 x^2 + 1)} + \frac{1}{(1 - 9 a) (a x^2 + 1)} \; \mathrm{d}x \\
&=\frac{\pi}{6} \left(\frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3}\right)\\
\end{align*}
Now, $I'(a)$ with respect to $a$ from $0$ to $4$:
\begin{align*}
I(4)&=\frac{\pi}{6} \int_0^4 \frac{\frac{19}{\sqrt{a}}+9}{3a+10\sqrt{a}+3} \; \mathrm{d}a \\
&=\frac{\pi}{6} \int_0^4 \frac{\frac{1}{\sqrt{a}}}{\sqrt{a}+3} + \frac{\frac{6}{\sqrt{a}}}{3\sqrt{a}+1} \; \mathrm{d} a \\
&=\frac{\pi}{6} \left(2 \int_0^4 \frac{\mathrm{d}\left(\sqrt{a}+3\right)}{\sqrt{a}+3} + 4 \int_0^4 \frac{\mathrm{d}\left(3\sqrt{a}+1\right)}{3\sqrt{a}+1}\right) \\
&=\frac{\pi}{3} \left(\ln{\left(\sqrt{a}+3\right)}+2\ln{\left(3\sqrt{a}+1\right)}\right) \bigg \rvert_0^4 \\
I(4) &= \int_{0}^{\infty}\frac{\log(1+4x^2)(1+9x^2)(9+x^2)+(9+x^2)\log(4+x^2)(10+10x^2)}{(9+x^2)^{2}(1+9x^2)} \mathrm{d}x =\boxed{\frac{\pi \ln{\left(\frac{245}{3}\right)}}{3}}\\
\end{align*}
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that a non-decreasing function $F$ with the properties $F(0) = 0$, $F(x/3)=F(x)/2$ and $F(1-x) = 1 - F(x)$ is the Cantor function $\begingroup$
I'm trying to prove by induction an assertion of Chalice (1991) article (http://people.math.sc.edu/girardi/m555/current/hw/FunkyFunctions/Cantor.pdf). I know that if $a_{n+1_{k}}$ is a $k$th bounder of an interval removed at the $(n+1)$th step of construction of the Cantor set $K$ or $a_{n+1_{k}}\in \{0,\ 1\}$, then $a_{n+1_{k}} \in \frac{1}{3}E_{n} \cup \left(\frac{2}{3}+\frac{1}{3}E_{n}\right)$, where $E_{n}$ is the set of the bounders of the removed intervals at the $n$th step of construction of $K$ in union with $\{0,\ 1\}$. If $a_{n+1_{k}}$ is at the first set, then I could prove by the following way:
$$a_{(n+1)_{k}} = \frac{1}{3}a_{n_{k}} \Rightarrow F\left(a_{(n+1)_{k}}\right) = F \left(\frac{1}{3}a_{n_{k}}\right)=\frac{1}{2}F(a_{n_{k}})$$
By the induction hypothesis:
$$ F\left(a_{(n+1)_{k}}\right) =\frac{1}{2} \mathcal{C} \left(a_{n_{k}}\right) = \frac{1}{2}\mathcal{C}\left(3(a_{(n+1)_{k}})\right)$$
where $\mathcal{C}$ is the Cantor function.
Just as $a_{(n+1)_{k}} \in E_{n+1} \cup \{0,\ 1\} \subset K $, so it can be written in its ternary expansion as
$$ a_{(n+1)_{k}} = \sum_{i=1}^{\infty} \frac{a_{(n+1)_{k_{i}}}}{3^{i}};\ a_{(n+1)_{k_{i}}} \in \{0,\ 2\} $$
So, by definition:
$$\mathcal{C} \left(3a_{(n+1)_{k}}\right) = \sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i}}}}{2^{i-1}}$$
and
\begin{align*} F(a_{(n+1)_{k}}) &= \frac{1}{2} \sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i}}}}{2^{i-1}}\\ &= \sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i}}}}{2^{i}} \\&= \mathcal{C} \left(a_{(n+1)_{k}}\right)
\end{align*}
just as desired.
Although, if $$a_{n+1_{k}} \in \left(\frac{2}{3}+\frac{1}{3}E_{n}\right)$$
I could go no further than \begin{align*}a_{(n+1)_{k}} = \frac{2}{3}+\frac{1}{3}a_{n_{k-\left(2^{n+1}\right)}} \\ &\Rightarrow F\left(a_{(n+1)_{k}}\right) = F\left(\frac{2}{3}+\frac{1}{3}a_{n_{k-\left(2^{n+1}\right)}}\right) \\ &\Rightarrow F(a_{(n+1)_{k}}) = 1-F\left(1-\frac{2}{3}-\frac{1}{3}a_{n_{k-\left(2^{n+1}\right)}}\right) = 1-F\left(\frac{1}{3}\left(1-a_{n_{k-\left(2^{n+1}\right)}}\right)\right) = 1-\frac{1}{2}F\left(1-a_{n_{k-\left(2^{n+1}\right)}}\right) \\ &\Rightarrow F(a_{(n+1)_{k}}) = 1-\frac{1}{2}\left(1-F\left(a_{n_{k-\left(2^{n+1}\right)}}\right)\right) = \frac{1}{2}+\frac{1}{2}F\left(a_{n_{k-\left(2^{n+1}\right)}}\right)\end{align*}
Now, by the induction hypothesis, $$F\left(a_{(n+1)_{k}}\right) = \frac{1}{2}+\frac{1}{2} \mathcal{C} \left(a_{n_{k-\left(2^{n+1}\right)}}\right) = \frac{1}{2}+\frac{1}{2} \mathcal{C} \left(3a_{(n+1)_{k}}-2\right)$$ That's the point I'm stuck, I'm not managing to prove that the right side of the last equation is $\mathcal{C} \left(a_{(n+1)_{k}}\right)$. Just explaining better the index $k$, I'm considering $a_{n_{0}} = 0 $ and so $0 \leq k \leq 2^{n+1}-1$ at th $n$th step of construction of $K$.
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I think I solved the step I was stucked on. Since $a_{n_{k-\left(2^{n+1}\right)}} \in E_{n} \cup \{0,\ 1\} \subset K $, we have $a_{(n+1)_{k}} = \frac{2}{3}+\frac{1}{3}a_{n_{k-\left(2^{n+1}\right)}} \Rightarrow a_{(n+1)_{k}} = \frac{2}{3}+\frac{1}{3}\sum_{i=1}^{\infty} \frac{a_{n_{k-\left(2^{n+1}\right)_{i}}}}{3^{i}} $, so, in base $3$, $a_{(n+1)_{k}} = 0,2a_{(n+1)_{k_{2}}}a_{(n+1)_{k_{3}}} \cdots $, with $a_{(n+1)_{k_{i}}} \in \{0,\ 2\} $. Then $ \left[3.a_{(n+1)_{k}}-2\right]_{10} = \left[10.a_{(n+1)_{k}}-2\right]_{3} = \left[2,a_{(n+1)_{k_{i+1}}}-2\right]_{3} = \sum_{i=1}^{\infty} \frac{a_{(n+1)_{k_{i+1}}}}{3^{i}}$.
By definition $ \frac{1}{2} + \frac{1}{2} \mathcal{C} \left(\sum_{i=1}^{\infty} \frac{a_{(n+1)_{k_{i+1}}}}{3^{i}}\right) = \frac{1}{2} + \left(\sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i+1}}}}{2^{i+1}}\right)$ and also $ \mathcal{C} (a_{(n+1)_{k}}) = \sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i}}}}{2^{i}}$. Since $ a_{(n+1)_{k_{1}}} = 2$, then $ \mathcal{C} (a_{(n+1)_{k}}) = \frac{2/2}{2^1}+\sum_{i=1}^{\infty} \frac{1}{2} \frac{a_{(n+1)_{k_{i+1}}}}{2^{i+1}} = \frac{1}{2} + \frac{1}{2} \mathcal{C}\left(3a_{(n+1)_{k}}-2\right) = F\left(a_{(n+1)_{k}}\right)$.
Now, $ F = \mathcal{C} $ in the removed intervals bounders, and both $ F $ and $ \mathcal{C} $ are constant (this result I've proved before) at each of these intervals in $[0,\ 1]/K$ then $ F = \mathcal{C} $ in $[0,\ 1]/K$. Going further, $ F = \mathcal{C} $ in the removed intervals bounders implies that they're equal in a dense subset of $K$, and so they're equal in each compact subset of $K$. Just as $K$ is compact, they're equal in $K$.
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|
Find the complete solution set of the equation ${\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is The complete solution set of the equation ${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) = \frac{\pi }{4} + {\sin ^{ - 1}}x$ is
(A)[-1,0]
(B)[0,1]
(C)$[-1,\frac{1}{\sqrt{2}}]$
(D)$[\frac{1}{\sqrt{2}},1]$
My approach is as follow
${\sin ^{ - 1}}\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right) - {\sin ^{ - 1}}x = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)\sqrt {1 - {x^2}} - x\sqrt {1 - {{\left( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \right)}^2}} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {1 - \left( {\frac{{{x^2} + 1 - {x^2} + 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
${\sin ^{ - 1}}\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{2 - 1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{\pi }{4}$
$\left( {\left( {\frac{{x\sqrt {1 - {x^2}} + 1 - {x^2}}}{{\sqrt 2 }}} \right) - x\sqrt {\left( {\frac{{1 - 2x\sqrt {1 - {x^2}} }}{2}} \right)} } \right) = \frac{1}{{\sqrt 2 }}$
Not able to proceed from here
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One way is using differentiation. Let $$f(x) = \sin^{-1}(x) + \frac{\pi}{4}$$ And $$g(x) = {\sin ^{ -1}}( {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} ) $$
Edit:
Note that domain of $\sin^{-1} x$ is $-1\le x \le 1$. The domain of $f(x) - g(x)$ is $A = \{x | x \in \mathbb{R}, x \in D_f \cap D_g \}$. So we have $$D_f = \{x | x \in \mathbb{R}, -1\le x \le 1\}$$And $$D_g = \{x | x \in \mathbb{R}, -1\le {\frac{{x + \sqrt {1 - {x^2}} }}{{\sqrt 2 }}} \le 1\}$$
If we show $x \in D_f \implies x \in D_g$ then we are done. Because it shows $D_f \subseteq D_g$ and then $D_f \cap D_g = D_f$. Consider $y = x + \sqrt {1 - {x^2}}$ on the interval $x \in [-1 , 1]$. It's a continuous function and $y(-1) =-1, \ \ y(1) = 1$. Taking derivative and finding critical points $$y'(x) = 1 - \frac{x}{\sqrt{1-x^2}} = 0 \implies x = \pm \frac{1}{\sqrt{2}}$$Because $y(\frac{1}{\sqrt{2}}) = \sqrt{2}$ and $y(-\frac{1}{\sqrt{2}}) = 0$ we have $$-1 \le x + \sqrt {1 - {x^2}} \le \sqrt{2} \implies \frac{-1}{\sqrt{2}} \le \frac{x + \sqrt {1 - {x^2}}}{\sqrt{2}} \le 1 $$ Since $-1 \le \frac{-1}{\sqrt{2}}$ the result is proved. This is a geometric interpretation for the mentioned inequity:
If $-1\lt x \lt 1$ then $$f'(x) -g'(x) = \frac{1}{\sqrt{1-x^2}} - \frac{\sqrt{1 - x^2} - x}{\sqrt{1 - 2x\sqrt{1 - x^2}} \sqrt{1 - x^2}} $$
Solve for $f'(x) - g'(x) = 0$:
$$\sqrt{1 - 2x\sqrt{1 - x^2}} = \sqrt{1 - x^2} - x $$
If $\sqrt{1 - x^2} - x \ge 0$ we can safely square both sides:
$$1 - 2x\sqrt{1 - x^2} = 1 - x^2 + x^2 - 2x\sqrt{1 - x^2} $$ Which is an identity. So if $-1\lt x \lt 1$ and $\sqrt{1 - x^2} - x \ge 0$ then $f'(x)- g'(x) = 0$. Therefore we have $f(x) = g(x) + C$. Also $f(0) = g(0) = \frac{\pi}{4}$ concluding $C = 0$. So we have to check the interval $$-1\lt x \lt 1 \ \ \ \text{and} \ \ \ \sqrt{1 - x^2} - x \ge 0$$
Which is $-1 \lt x \le \frac{1}{\sqrt{2}}$. You can easily check $x = -1$ by hand.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Convergence or divergence of $\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$ using integral test
Finding convergence or Divergence of series $$\sum^{\infty}_{n=1}\frac{2^n\cdot n^5}{n!}$$ using integral test.
What i try::
$$\frac{2}{1}+\frac{4\cdot 32}{2!}+\frac{8\cdot 3^5}{3!}+\sum^{\infty}_{n=4}\frac{2^n\cdot n^5}{n!}$$
Using $$\frac{n!}{2^n}=\frac{4!}{2^4}\frac{5}{2}\frac{6}{2}\cdots \cdots >1\Longrightarrow \frac{n!}{2^n}>1$$
So $$\frac{2^n}{n!}\leq 1\Longrightarrow \frac{2^n\cdot n^5}{n!}<n^5$$
How do i solve it Help me please, Thanks
|
The integral test requires an upper bound which is not immediate. Instead we can relate the series to the Taylor expansion of the exponential.
$$n^5=n(n-1)(n-2)(n-3)(n-4)+10(n-1)(n-2)(n-3)(n-4)\\+65(n-2)(n-3)(n-4)+285(n-3)(n-4)+781(n-4)+1024$$ so that
$$\sum_{n=1}^\infty\frac{2^nn^5}{n!}=
\\\sum_{n=1}^\infty2^n\left(\frac1{(n-5)!}+\frac{10}{(n-4)!}+\frac{65}{(n-3)!}+\frac{285}{(n-2)!}+\frac{781}{(n-1)!}+\frac{1024}{n!}\right)$$ (the terms with a factorial of a negative must be omitted).
This sum is obviously convergent and equals $ae^2+b$ where $a,b$ are rational numbers.
[You can show convergence in an easier way (see @zkutch); but my goal was to show that one can find the exact value.]
|
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|
Conditional Probability Denominator doubt Let X be a rv such that $P(X = 2) = 1/4$ and its CDF is given by
$$F_{X}(x) =
\begin{cases}
0,& x< -3\\
\frac{3}{4}(x+3),& -3\leq x <2 \\
3/4,& 2 \leq x <4\\
\frac{3}{64} x^2,& 4 \leq x < \frac{8}{\sqrt{3}} \\
1,& x \geq \frac{8}{\sqrt{3}}
\end{cases}
$$
$x=2$ in the only discontinuity of $F$.
I have to compute $P(X<3|X \geq 2)$.
My way is: $\frac{P(2 \leq X<3)}{P(X \geq2)}$ = $\frac{F(3)- F(2) - P(X = 3) + P(X=2)}{1 - P(X<2)}$
$P(X= 3) = 0$ because it is continuous there.
How can I compute the denominator.
|
$P(X>2)=\int_4^{8/\sqrt3 } \frac 3 {32} xdx=\frac 1 4$. Add $P(X=2)=\frac 1 4$ to this to get $P(X \geq 2)$.
|
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|
How do I find integers $x,y,z$ such that $x+y=1-z$ and $x^3+y^3=1-z^2$? This is INMO 2000 Problem 2.
Solve for integers $x,y,z$: \begin{align}x + y &= 1 - z \\ x^3 + y^3 &= 1 - z^2 . \end{align}
My Progress: A bit of calculation and we get $x^2-xy+y^2=1+z $
Also we have $x^2+2xy+y^2=(1-z)^2 \implies 3xy=(1-z)^2-(1+z)=z(z-3) \implies y=\frac{z(z-3)}{3x}$ and $x=\frac{z(z-3)}{3y} $.
Note that since $z$,$x$,$y$ is an integer, we must have $3\mid z$.
So, let $z=3k$.
So we have $y=\frac{3k(3k-3)}{3x}=\frac{k(3k-3)}{x}$ and $x=\frac{z(z-3)}{3y}=\frac{k(3k-3)}{y}$ .
Then I am not able to proceed.
Hope one can give me some hints and guide me.
Thanks in advance.
|
A general way:
By elimination of $z$ we get
$$(x+y)(x^2+y^2-xy+x+y-2)=0$$
Case 1: So two branches one: $x+y=0 \implies z=1,x=n, y=-n$, where $n\in I$
Case 2: The other set of solutions are given by
$$(x^2+y^2+xy+x+y-2)=0$$, write this as a quadratic of $x$ and treat $y$ as constant then
$$x=\frac{-(1-y)\pm \sqrt{(1-y)^2-4(y^2+y-2)}}{2}$$
$$x=\frac{y-1\pm\sqrt{-3[(y-3)(y+1)}}{2}~~~~(1)$$
The reality demands that $-3(y-3)(y+1)\ge 0 \implies (y-3)(y+1)\le 0$
\implies that $-1\le y\le 3$. So the possible integral values of $y$ are: $-1, 0, 1,2,3$ out of these only $y=1$ gives $x$ as irrational.
$$y=-1 \implies x=-1, y=0 \implies x=1,-2; y=2 \implies x=-1,2; y=3 \implies x=1$$
We get six integral pairs of $(x,y)$, for them $z=1-x-y$ will yield six triplets of $(x,y,z)$
|
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|
Compute the matrix of norms of $A=\begin{bmatrix}3&4\\1&-3\end{bmatrix}$ My work so far
Using the following
$\hspace{30px} L^1\ =\displaystyle \max_{\small 1\le j\le m}(\displaystyle \sum_{i=1}^n |a_{ij}|)\\ \hspace{30px} L^2\ =\sigma_{max}(A)\\ \hspace{30px} L^F\ =\sqrt{\displaystyle \sum_{i} \displaystyle \sum_{j} |a_{ij}|^2}\\ \hspace{30px} L^\infty\ =\displaystyle \max_{\small 1\le i\le n}(\displaystyle \sum_{j=1}^m |a_{ij}|)\\$
Thus,
$L^1=\begin{bmatrix}3&\textbf{4}\\1&\textbf{3}\end{bmatrix}=7\\
L^2=?\\
L^F=\sqrt{3^2+4^2+1^2+(-3)^2}=\sqrt{35}=5.916079783\\
L^\infty=\begin{bmatrix}\textbf{3}&\textbf{4}\\1&-3\end{bmatrix}=7$
However, I'm unsure how to get $L^2$. How would I start off doing this part?
|
Alternatively, recall that
$$\|A\|_2 = \sup_{\|(x,y)\|_2=1} \|A(x,y)\|_2 = \sup \left\{\left\|\begin{bmatrix}3&4\\1&-3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\right\| : x^2+y^2=1\right\}$$
so subject to $x^2+y^2=1$ we wish to maximize $$f(x,y) = (3x+4y)^2+(x-3y)^2 = 10x^2+25y^2+18xy.$$
Plugging in $y=\sqrt{1-x^2}$ this is $$g(y)=10+15y^2+18y\sqrt{1-y^2}.$$
Solving $g'(y) = 0$ for $y\in[0,1]$ gives $y=\sqrt{\frac12+\frac{5}{2\sqrt{61}}}$ and finally
$$g(y) = \frac{35+ 3\sqrt{61}}{2}$$
so $$\|A\|_2 = \sqrt{g(y)} = \sqrt{\frac{35+ 3\sqrt{61}}{2}}.$$
|
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|
Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
|
the pqr method:
Let $p = a+b+c, q = ab+bc+ca = 3, r = abc$.
Since $(a+b+c)^2 - 3(ab+bc+ca) = a^2+b^2+c^2 - ab-bc-ca \ge 0$, we have $p \ge 3$.
Since $ab+bc+ca \ge 3\sqrt[3]{ab \cdot bc \cdot ca} = 3\sqrt[3]{(abc)^2}$, we have $r\le 1$.
We have
\begin{align}
(1+a^2)(1+b^2)(1+c^2) - 8 &= a^2b^2c^2+a^2b^2+b^2c^2+c^2a^2+a^2+b^2+c^2-7\\
&= r^2 + q^2 - 2pr + p^2-2q - 7 \\
&= (p-r)^2 - 4 \\
&\ge (3-1)^2 - 4\\
&= 0.
\end{align}
We are done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$ Solve $\sqrt[4]{x}+\sqrt[4]{x+1}=\sqrt[4]{2x+1}$
My attempt:
Square both sides three times
$$\begin{align*}
36(x^2+x)&=4(\sqrt{x^2+x})(2x+1+\sqrt{x^2+x})\\
(\sqrt{x^2+x})(35\sqrt{x^2+x}-4(2x+1))&=0
\end{align*}$$
This means $0,-1$ are solutions but I can't make sure that these are the only solutions. Also I'm not sure that squaring three times is a good approach or not.
|
Let $\sqrt[4]{x}=a$ and $\sqrt[4]{x+1}=b$.
Thus, $a\geq0,$ $b\geq1$, $b^4-a^4=1$ and $$a+b=\sqrt[4]{a^4+b^4}$$ or
$$(a+b)^4=a^4+b^4$$ or $$2ab(2a^2+3ab+2b^2)=0,$$ which gives $$ab=0.$$
Can you end it now.
|
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|
How to prove that $S=\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)?$ How to prove that
$$S=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}=\frac{\pi^2}{16}-\frac{1}{4}\log^2(\sqrt{2}-1)$$
My attempt:
We have for $|x|\leq1$ $$\tanh^{-1}(x)=\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}$$
and :
\begin{align*}
\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\ \mathrm{d}x&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{x}\displaystyle\sum_{n=0}^{\infty}\frac{x^{2n+1}}{2n+1}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{1}{2n+1}\displaystyle\int_0^{\sqrt{2}-1}x^{2n}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=S\\
\end{align*}
So :
\begin{align*}
S&=\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{2n+1}}{(2n+1)^2}\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\tanh^{-1}(x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}\frac{1}{2x}\left(\log(1+x)-\log(1-x)\right)\mathrm{d}x\\
&=\frac{1}{2}(J_1-J_2)
\end{align*}
Where:
\begin{align*}
J_1&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1+x)}{x}\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}\displaystyle\int_0^{\sqrt{2}-1}x^n\mathrm{d}x\\
&=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
And:
\begin{align*}
J_2&=\displaystyle\int_0^{\sqrt{2}-1}\frac{\log(1-x)}{x}\mathrm{d}x\\
&=\displaystyle\int_0^{\sqrt{2}-1}-\displaystyle\sum_{n=0}^{\infty}\frac{x^n}{n+1}dx\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{1}{n+1}\left[\frac{x^{n+1}}{n+1}\right]_0^{\sqrt{2}-1}\\
&=-\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}}{(n+1)^2}\\
\end{align*}
Finally we find :
$$S=\frac{1}{2}\left(\displaystyle\sum_{n=0}^{\infty}\frac{(\sqrt{2}-1)^{n+1}((-1)^n+1)}{(n+1)^2}\right)$$
But I could not find a way to calculate $S$.
Any help please? and thank's in advance.
|
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\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
S & \equiv \sum_{n = 0}^{\infty}{\pars{\root{2} - 1}^{2n + 1} \over
\pars{2n + 1}^{2}} =
\sum_{n = 1}^{\infty}{\pars{\root{2} - 1}^{n} \over
n^{2}}\,{1^{n} - \pars{-1}^{n} \over 2}
\\[5mm] & =
{1 \over 2}\sum_{n = 1}^{\infty}{\pars{\root{2} - 1}^{n} \over n^{2}} -
{1 \over 2}\sum_{n = 1}^{\infty}{\pars{1 - \root{2}}^{n} \over n^{2}}
\\[5mm] & =
{\mrm{Li}_{2}\pars{\root{2} - 1} - \mrm{Li}_{2}\pars{1 - \root{2}}
\over 2}
\approx 0.4226
\end{align}
$\ds{\mrm{Li}_{s}}$ is the Polylogarithm function.
$\color{red}{\tt\large See}$ the nice ${\tt @Naren}$ answer who already worked out the details toward the final answer
$\ds{\bbox[10px,#ffd,border:2px groove navy]{{\pi^{2} \over 16}
- {1 \over 4}\,\ln^{2}\pars{\root{2} - 1}}}$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof that $ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $
If
$$2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \hspace{1cm} \forall x \in \mathbb{R},$$
Proof that:
$$ \frac{2^{-x}-1}{x} = \sum_{n=0}^{\infty} \frac{ (-1)^{n+1}x^n(\ln2)^{n+1}}{(n+1)!} $$
I did the following:
\begin{align*}
&2^x = \sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!} \\
\Rightarrow \quad & 2^{-x} = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} \\
\Rightarrow \quad & 2^{-x}-1 = \frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1 \\
\Rightarrow \quad & \dfrac{2^{-x}-1}{x} = \frac{\frac{1}{\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}} -1}{x} \\
\Rightarrow \quad & \frac{2^{-x}-1}{x} = \frac{1-\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}{x\sum_{n=0}^{\infty} \frac{(x\ln(2))^n}{n!}}
\end{align*}
From this part I don't know how to continue
|
Your series can be written:
$$\sum_{n=0}^{\infty} \frac{\left(-1\right)^{n + 1} x^{n} \ln{\left(2 \right)}^{n + 1}}{\left(n + 1\right)!}=\sum_{n=0}^{\infty} \frac{x^{n} \left(- \ln{\left(2 \right)}\right)^{n + 1}}{\left(n + 1\right)!}$$
being $(-1)^{n+1}=-1$ forall $n\in\Bbb N$.
And rewriting
$$\sum_{n=0}^{\infty} \frac{x^{n} \left(- \ln{\left(2 \right)}\right)^{n + 1}}{\left(n + 1\right)!}=\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n + 1}}{x \left(n + 1\right)!}$$
Shift the series by $1$ you obtain:
$$\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n + 1}}{x \left(n + 1\right)!}=\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{x n!}=\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{x n!}= \frac{\displaystyle\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}}{x}$$
(remember that $x$ it is a constant).
After,
$$\frac{\sum_{n=1}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}}x=\frac{\left(\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!} + \sum_{n=0}^{0} - \frac{\left(- x \ln{\left(2\right)}\right)^{n}}{n!}\right)}{x}=\frac{\sum_{n=0}^{\infty} \frac{\left(- x \ln{\left(2 \right)}\right)^{n}}{n!}-1}{x}=\frac{e^{- x \ln{\left(2 \right)}}-1}{x}$$
(you have an exponential series)
i.e.
$$\frac{e^{- x \ln{\left(2 \right)}}-1}{x}=\frac{2^{-x}-1}{x}$$
|
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|
Bernoulli's Inequality for $-1 \leq x\leq 0$ My original goal was to prove that
$$\lim_{x\to 0}\frac{e^x-1}{x}=1$$
using the squeeze theorem as we haven't seen differentiability yet and thus I cannot use arguments such as Taylor series nor Bernoulli's theorem, nor can I use induction. For that I wanted to find a lower and upper bound for $e^x$ in order to apply the squeeze theorem.
For the upper bound I used the fact that $x^n\leq x^2$ for $-1\leq x\leq 1$ and $n\geq 2$ thus one has that
\begin{align*}e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n&=\lim_{n\to\infty}\sum_{k=0}^n{n\choose k}\frac{x^k}{n^k}\\
&=\lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\\
&\leq \lim_{n\to\infty}1+x+\sum_{k=2}^n{n\choose k}\frac{x^2}{n^k}\\
&= \lim_{n\to\infty}1+x+\left(\sum_{k=2}^n{n\choose k}\frac{1}{n^k}\right)\cdot x^2\\
&= \lim_{n\to\infty}1+x+\left(\left(1+\frac{1}{n}\right)^n-2\right)\cdot x^2\\
&= 1+x+\left(e-2\right)\cdot x^2
\end{align*}
I could now potentially bound $x^n\geq -x^2$ in the same interval and obtain the bound
\begin{align*}e^x\geq 1+x-\left(e-2\right)\cdot x^2
\end{align*}
but I am not happy with it as I know that Bernoulli's inequality is stronger and gives
\begin{align*}e^x\geq 1+x.
\end{align*}
For $x\in (0,1)$ it's rather trivial to prove as
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+\underbrace{{n\choose 2}\frac{x^2}{n^2}+\cdots+\frac{x^n}{n^n}}_{\geq 0}\geq 1+x$$
but for $x\in(-1,0)$ the same argument does not apply straightforwardly due to the changing signs. So I modified it as follows:
For $-1\leq x\leq 0$ one has that $x^3\leq x^n$ ($x^3$ is in particular negative)
\begin{align*}
e^x&=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n=\lim_{n\to\infty}1+x+{n\choose 2}\frac{x^2}{n^2}+\sum_{k=3}^n{n\choose k}\frac{x^k}{n^k}\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\sum_{k=3}^n{n\choose k}\frac{x^3}{n^k}\\
& = \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left((1+\frac{1}{n})^n-\frac{n-1}{2n}-2\right)x^3\\
& \geq \lim_{n\to\infty}1+x+\frac{(n-1)x^2}{2n}+\left(e-\frac{n-1}{2n}-2\right)x^3\\
&= \lim_{n\to\infty}1+x+x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)
\end{align*}
Now we not that the cubic function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ has a double zero at $x=0$ and the remaining zero is at
$$x=-\frac{\frac{n-1}{2n}}{e-\frac{n-1}{2n}-2}\overset{n\to \infty}{\longrightarrow} -\frac{1/2}{e-1/2-2}\cong -2.29$$
thus for $n$ sufficiently large the last zero is to the left of $-1$ and hence the function $x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)$ is positive on $(-1,0)$ thus
\begin{align*}
e^x& \geq \lim_{n\to\infty}1+x+\underbrace{x^2\left(\frac{n-1}{2n}+\left(e-\frac{n-1}{2n}-2\right)x\right)}_{\geq 0,\quad x\in(-1,0)}\\
&\geq 1+x
\end{align*}
Since I wrote up this proof I ask: could please give it a look and tell me if there are any mistakes or if there is a shorter solution which I overlooked?
Many thanks in advance!
|
As regards your original goal, there is a shorter way. We have that
$$e^x-1-x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n-1-x=\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}.$$
Hence, for $x\in [-1,1]$, $$|e^x-1-x|=\left|\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{x^k}{n^k}\right|\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{|x|^{k-2}}{n^k}\\\leq x^2\lim_{n\to\infty}\sum_{k=2}^n{n\choose k}\frac{1}{n^k}
\leq x^2\sum_{k=2}^{\infty}\frac{1}{k!}\leq e x^2$$
and the given limit follows as $x\to 0$ by the squeeze theorem.
Along the same argument we show that for $x\in [-1,1]$,
$$\left|e^x-\sum_{k=0}^{n}\frac{x^{k}}{k!}\right|<e|x|^{n+1}$$
which implies that $e^x=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}$.
Bernoulli inequality for $-1<x<0$:
$$e^x-1-x=\sum_{k=2}^{\infty}\frac{x^{k}}{k!}=\sum_{k=1}^{\infty}\frac{x^{2k}}{(2k)!}\underbrace{\left(1+\frac{x}{2k+1}\right)}_{\geq 0}\geq 0.$$
|
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|
Counting the number of integers with given restrictions Question: Consider the numbers $1$ through $99,999$ in their ordinary decimal representations. How many contain exactly one of each of the digits $2, 3, 4, 5$?
Answer: $720$.
Attempt at deriving the answer:
We have two cases: four digit numbers and five digit numbers.
Five digit numbers:
Let $x \in \{2, 3, 4, 5\}$. If the first position in a five digit number is not $x$, then there are $5$ possibilities for this position as there are four values for $x$ and $0$ is inadmissable. The rest of the four positions will have the various permutations of four values of $x$. There are $5 \times 4!$ such numbers. If the non-$x$ value is in the second position, then there are $4$ ways to choose an $x$-value for the first position, $6$ integers for the second position and $3!$ permutations for the rest of the positions. There are $4\times 6 \times 3!$ such numbers. If the non-$x$ value is in the third position, then there are $\binom 42$ ways to choose two $x$-values, $2!$ ways to permute them and $6$ integers for the third position meaning there are $6 \times 2 \times 6$ such numbers. When the non-$x$ value is in the fourth position, there are $4 \times 3! \times 6$ such numbers. Finally, if non-$x$ is in the fifth position, there are $4! \times 6$ such numbers.
Four digit numbers:
We just need to permute the number $2345$. There are $4!$ such permutations.
Thus the number of numbers with the given restrictions is $5\times 4! + 4\times 3!\times 6 + 2\times 6 \times 6 + 4 \times 3! \times 6 + 6 \times 4! + 4! = 648$.
What did I forget to take into account? Thanks.
|
Write $0$ at the front of four digits number so we always have five digits number. The fifth number is one of $(0,1,6,7,8,9)$.
$6\times 5! =720$
|
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|
(BAMO $2013/3$) $ABH$, $BCH$ and $CAH$ is congruent to $ABC$.
Let $H$ be the orthocenter of an acute triangle $ABC$. Prove that the triangle formed by the circumcenters of triangles $ABH$, $BCH$ and $CAH$ is congruent to $ABC$.
I have already seen many answers on MSe But my doubt is diffferent,
In this solution (first one )
https://artofproblemsolving.com/community/c618937h1628954_problem_320_bamo_20133
To prove $O_AB || O_BA$, we can do some angle calculations: $\angle O_ABC = 90 - A$, and $\angle C A O_B = 90 - B$
how he got $\angle O_ABC = 90 - A$, and $\angle C A O_B = 90 - B$ ?
I tried some angle chasing but did not able to get this ..thankyou
|
Because $AO_B=CO_B$, $\measuredangle AO_BC=2\beta$ and from here:
$$\measuredangle CAO_B=90^{\circ}-\beta.$$
Another way.
Let $A'$, $B'$ and $C'$ be circumcenters of $\Delta BHC$, $\Delta CHA$ and $\Delta AHB$ respectively.
Thus, since $B'HA'C$ is a rhombus and $HC=c\cot\gamma$ in the standard notation, we obtain:
$$A'B'=2\sqrt{B'H^2-\left(\frac{HC}{2}\right)^2}=2\sqrt{R^2-\frac{1}{4}c^2\cot^2\gamma}=$$
$$=2\sqrt{\frac{a^2b^2c^2}{16S^2}-\frac{1}{4}c^2\cdot\frac{\left(\frac{a^2+b^2-c^2}{2ab}\right)^2}{\frac{4S^2}{a^2b^2}}}=2\sqrt{\frac{a^2b^2c^2}{16S^2}-\frac{c^2(a^2+b^2-c^2)^2}{64S^2}}=$$
$$=\frac{c}{4S}\cdot\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}=\frac{c}{4S}\cdot\sqrt{\sum_{cyc}(2a^2b^2-a^4)}=c.$$
Similarly $A'C'=b$ and $B'C'=a$ and we are done!
|
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|
Find an approximation to $\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor (j-1)$ I want to find an approximate (ideally asymptotic) function $f_1:\mathbb{R}\to\mathbb{R}$ in order to approximate a function $f_0:\mathbb{R}\to\mathbb{N}$ with $f_0$ defined by
$$\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor (j-1)$$
My attempts so far have not worked, and I'd really appreciate a workable solution.
I know the following is wrong, and contains several unjustified steps. What I am after is primarily a solution that works, but it would of course also be good to see where I went wrong.
My first attempt was to assume that for large $x$, $\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor (j-1)\approx\sum _{j=1}^x (\frac{x}{j} -1)(j-1)$, giving
$$\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor(j-1)\approx\sum _{j=1}^x (\frac{x}{j} -1) (j-1)$$$$=\frac{1}{2}(x^2+x-2 H_x)$$
where $H_n$ is the $n$th harmonic number. However, the following plot suggests (heuristically) that the ratio
$$\frac{\frac{1}{2}(x^2+x-2 H_x)}{\sum _{j=1}^x (\frac{x}{j} -1)(j-1)}$$
has an asymptote at a value slightly greater than $3$:
Presumably, I could therefore get a reasonable approximation by writing
$$\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor(j-1)\approx\frac{1}{6}(x^2+x-2 H_x)$$
but I would have no logical justification for the approximation.
For my second attempt, I reasoned that in any half-open interval $a=(b,b+1]$ with $b$ a positive integer, the 'average' value (if 'average' is the right term) of $\left\lfloor a\right\rfloor$ is $b$, whereas the average value of the real number $a$ is $\frac{2b+1}{2}$ - and therefore the average value of the ratio $\frac{a}{\left\lfloor a\right\rfloor}\approx\frac{2b+1}{2b}$. I then assumed that this could be extended from integer $b$ to real $a$, giving the rough equivalence $\frac{a}{\left\lfloor a\right\rfloor}\approx\frac{2a+1}{2a}$. (This is almost certainly wrong!)
I then substituted $a\to(\frac{x}{j}-1)$ and wrote $$\frac{\frac{x}{j}-1}{\left\lfloor \frac{x}{j}-1\right\rfloor}\approx\frac{2(\frac{x}{j}-1)+1}{2(\frac{x}{j}-1)}$$ $$\implies\left\lfloor \frac{x}{j} -1\right\rfloor\approx \frac{2}{2(\frac{x}{j}-1)+1}$$
But, heuristically, the ratio
$$\frac{\sum _{j=1}^x \frac{2}{2(\frac{x}{j}-1)+1}(j-1)}{\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor(j-1)}$$
appears to converge (more slowly than my first attempt) towards a value somewhat less than $5$:
So: two failed attempts...
How do I go about finding a valid approximation to $\sum _{j=1}^x \left\lfloor \frac{x}{j} -1\right\rfloor (j-1)$?
|
I suppose that $f_0(x) = 0$ for $x < 1$. Otherwise it's not clear how the sum should be interpreted, but with reasonable interpretations the behaviour can be determined much like the behaviour for $x \geqslant 1$.
First we split the terms $\bigl\lfloor \frac{x}{j} - 1\bigr\rfloor (j-1) = \bigl\lfloor \frac{x}{j}\bigr\rfloor j - \bigl\lfloor\frac{x}{j}\bigr\rfloor - (j-1)$.
For the sum over the first parts of the split terms we obtain
\begin{align}
\sum_{j \leqslant x} \biggl\lfloor \frac{x}{j}\biggr\rfloor j
&= \sum_{j \leqslant x} j\sum_{k \leqslant x/j} 1 \\
&= \sum_{k\cdot j \leqslant x} j \\
&= \sum_{k \leqslant x} \sum_{j \leqslant x/k} j \\
&= \frac{1}{2} \sum_{k \leqslant x} \biggl\lfloor \frac{x}{k}\biggr\rfloor \Biggl(\biggl\lfloor \frac{x}{k}\biggr\rfloor + 1\Biggr) \\
&= \frac{1}{2} \sum_{k \leqslant x} \biggl(\frac{x^2}{k^2} + O\biggl(\frac{x}{k}\biggr)\biggr) \\
&= \frac{x^2}{2}\biggl(\frac{\pi^2}{6} + O\biggl(\frac{1}{x}\biggr)\biggr) + O(x\log x) \\
&= \frac{\pi^2}{12}x^2 + O(x\log x)\,.
\end{align}
The error term in this cannot be much improved, but a little. Walfisz proved an $O(x(\log x)^{2/3})$ bound, and since $\limsup \frac{\sigma(n)}{n\log \log n} = e^{\gamma}$, the error term cannot be smaller than $O(x\log \log x)$.
If we use the error term derived above, we can completely ignore the second parts, since their sum is $O(x\log x)$, hence swallowed by the error term. If we use the stronger bound by Walfisz, the easy
$$\sum_{j \leqslant x} \biggl\lfloor \frac{x}{j}\biggr\rfloor = \sum_{j \leqslant x} \frac{x}{j} + O(x) = x\log x + O(x)$$
is all that we can fruitfully use. The stronger results for the summatory divisor function by Dirichlet and later improvements help not at all, everything but the main term is necessarily absorbed by the error term from the first parts.
Finally we know
$$\sum_{j \leqslant x} (j-1) = \frac{\lfloor x\rfloor(\lfloor x\rfloor - 1)}{2} = \frac{x^2}{2} + O(x)\,.$$
Putting everything together we obtain
$$f_0(x) = \frac{\pi^2}{12} x^2 - x\log x - \frac{1}{2}x^2 + O\bigl(x(\log x)^{2/3}\bigr) = \frac{\pi^2 - 6}{12}x^2 - x\log x + O\bigl(x(\log x)^{2/3}\bigr)$$
using Walfisz's result, and
$$f_0(x) = \frac{\pi^2-6}{12} x^2 + O(x\log x)$$
without appealing to that.
|
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|
How to find the side of the square ,by using trigonometry There is a square with diagonal length of 'a", the question is to find the length of the sides. it can be found by Pythagorean theorem. but I tried to do it with trigonometry, considering the properties of a square but it doesn't resemble the first answer, so I was wondering what did I forget/mistake?
|
Let the side length of the square equal $b$, and the hypotenuse $a$. By Pythagoras' Theorem, we have
\begin{align}
b^2+b^2&=a^2 \\
2b^2&=a^2 \\
b^2&=\frac{a^2}{2}\\
b&=\sqrt{\frac{a^2}{2}}=\frac{\sqrt{a^2}}{\sqrt 2}=\frac{a}{\sqrt 2}=\frac{\sqrt 2}{2}a
\end{align}
Alternatively, the angle between $a$ and $b$ is $45$ degrees. We know this because a square has $4$ right angles, which are bisected (cut in half) by the diagonal. Therefore,
\begin{align}
\sin(45)&=\frac{b}{a}\\
\end{align}
Now find the value of $\sin(45)$, and rearrange the equation to find $b$. Hopefully this answer is familiar.
It seems that part of your confusion stems from the fact that you are unsure about rationalising the denominator:
$$
\frac{1}{\sqrt 2}=\frac{\sqrt 2}{2}
$$
Consider the left-hand side of the above equality. Multiplying $\frac{1}{\sqrt 2}$ by $1$ does not change its value:
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}
$$
We also know that
$$
1=\frac{\sqrt{2}}{\sqrt{2}}
$$
Therefore,
$$
\frac{1}{\sqrt 2}\times1=\frac{1}{\sqrt 2}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt 2}{2}
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
The value of $\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $ is $$\int_0^1\ (\prod_{r=1}^n x+r).(\sum_{k=1}^n \frac{1}{x+k}).dx $$
My working:
$$\int_0^1\ [(x+1)(x+2)(x+3)...(x+n)]\times[\frac{1}{x+1}\ + \frac{1}{x+2}\ + \frac{1}{x+3}\ +\ ...\ + \frac{1}{x+n}].dx \\$$
$\int_0^1[(x+1)(x+2)(x+3)...(x+n)]\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{(x+1)(x+2)(x+3)...(x+n)}.dx\\
$
Now the numerator of the left expression and the denominator of the right expression will cancel out $\\$
$\int_0^1\require{enclose}\enclose{downdiagonalstrike}{[(x+1)(x+2)(x+3)...(x+n)]}\ \times\ \frac{[(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+n-1)]}{\enclose{downdiagonalstrike}{(x+1)(x+2)(x+3)...(x+n)}}.dx$
Now we are left with
$$\int_0^1 [(x+2)(x+3)..(x+n)\ +\ (x+1)(x+3)..(x+n)\ +\ ...\ +\ (x+1)(x+2)..(x+(n-1))].dx$$
I am unable to think of anything after this step although, I have an intuition that it may have something to do with factorials.
|
The integrand function is the derivative of $(x+1)(x+2)\cdots(x+n)=(x+1)_n$, hence the outcome simply is
$$\left[(x+1)_n\right]_{0}^{1}=(n+1)!-n!=n\cdot n!.$$
|
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|
Find limit of $f(x)$ as $x$ tends to $0$ I need some help answering this question:
$$f(x) = \frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}$$
find the limit of $f(x)$ as $x$ tends to $0$ by writing $f(x)$ as a quotient of two powers series.
I have so far:
$$\frac{(x(1+\frac{x^2}{2!}+\frac{x^4}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{x(x + \frac{x^3}{3!}+\cdots)}= \frac{(x+\frac{x^3}{2!}+\frac{x^5}{4!}+\cdots))-(x + \frac{x^3}{3!} + \cdots)}{(x^2 + \frac{x^4}{3!}+\cdots)}$$
but I don't know how to reduce this further.
|
By the definition of hyperbolic functions and $e^x=1+x+\frac{x^2}2+o(x^2)$ we have
$$\frac{\cosh(x)}{\sinh(x)} - \frac{1}{x}=\frac{e^{2x}+1}{e^{2x}-1} - \frac{1}{x}=\\=\frac{2+2x+2x^2+o(x^2)}{2x+2x^2+o(x^2)} - \frac{1}{x}=\frac{2x+2x^2-2x-2x^2+o(x^2)}{2x^2+o(x^2)}=\frac{o(1)}{2+o(1)} \to 0$$
|
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|
Product/division of exponencials with factorials as the base I put this formula on WolframAlpha $$\frac{(26!)^{n+2}}{13!}$$ and it simplified to $$2^{23n+36}\cdot175^{3n+5}\cdot7429^{n+2}\cdot34749^{2n+3}$$
I tried solving it by hand
\begin{align}
\frac{(26!)^{n+2}}{13!} & = \frac{(26\cdot25\cdot\ldots\cdot3\cdot2)^{n+2}}{13\cdot12\cdot\ldots\cdot3\cdot2} \\\\
& = (26\cdot25\cdot\ldots\cdot15\cdot14)^{n+2} \cdot (13\cdot12\cdot\ldots\cdot3\cdot2)^{n+1} \\\\
& = (2^{13}\cdot3^5\cdot5^4\cdot7^2\cdot11\cdot13\cdot17\cdot19\cdot23)^{n+2}\cdot(2^{10}\cdot3^5\cdot5^2\cdot7\cdot11\cdot13)^{n+1} \\\\
& = 2^{23n+36}\cdot(3^5)^{2n+3}\cdot(5^2)^{3n+5}\cdot7^{3n+5}\cdot11^{2n+3}\cdot13^{2n+3}\cdot17^{n+2}\cdot19^{n+2}\cdot23^{n+2} \\\\
& = 2^{23n+36}\cdot(5^2\cdot7)^{3n+5}\cdot(17\cdot19\cdot23)^{n+2}\cdot(3^5\cdot11\cdot13)^{2n+3}\\\\
& = 2^{23n+36}\cdot175^{3n+5}\cdot7429^{n+2}\cdot34749^{2n+3}\\
\end{align}
My question is do I have to do all this work (prime factorization of factorials) everytime I want to simplify an expression of this kind or is there a faster way using Discrete Mathematics?
|
For more systematic approach, let $v_p(x)$ be a $p$-adic valuation of $x$, then by some basic rules such as $v_p(x^n)=n\cdot v_p(x)$ and $v_p(x/y)=v_p(x)-v_p(y)$. Also we can use Legendre's formula $v_p(n!)=\sum_{i=1}^{\infty}\lfloor \frac{n}{p^i} \rfloor$, so
\begin{align}
v_2\left(\frac{(26!)^{n+2}}{13!}\right)&=(n+2)v_2(26!)-v_2(13!)\\
&=(n+2)\left(\left\lfloor \frac{26}{2} \right\rfloor + \left\lfloor \frac{26}{4} \right\rfloor + \left\lfloor \frac{26}{8} \right\rfloor + \left\lfloor \frac{26}{16} \right\rfloor \right)\\
&\ \ \ -\left(\left\lfloor \frac{13}{2} \right\rfloor + \left\lfloor \frac{13}{4} \right\rfloor + \left\lfloor \frac{13}{8} \right\rfloor \right)\\
&=(n+2)(13+6+3+1)-(6+3+1)\\
&=23n+36.\\
\end{align}
Now just continue in similar manner for $p=3,5,7,11,13,17,19,23$ to obtain complete factorization. Then you can group the terms together in any way you like.
|
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|
Why does $-8^{\frac{1}{3}}$ have $2$, $e^{\frac{\pi}{3}}$ and $e^{\frac{5\pi}{3}}$? Use DeMoivre’s theorem to find $-8^{\frac{1}{3}}$. Express your answer in complex form.
Select one:
a. –2
b. – 2, 2 cis ($\pi$/3)
c. – 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
d. 2, 2 cis ($\pi$/3), 2 cis (5$\pi$/3)
e. None of these
The correct answer is : $2, 2 e^{\pi/3}, 2 e^{5\pi/3}$
My calculation is here:
$r=\sqrt{-8^{2}}=8$
Then,
$= 2\ cis\ \frac{2\pi k}{n}$
If k is $0$,
$= 2\ cis\ 0=2$
If k is $1$,
$= 2\ cis\ \frac{\pi }{3}$
If k is $2$,
$= 2\ cis\ \frac{4\pi }{3}$
Therefore, the results are $2$, $= 2\ cis\ \frac{\pi }{3}$, and $= 2\ cis\ \frac{4\pi }{3}$.
So, why the correct answer is $2$, $2\ cis (\frac{\pi}{3})$, $2\ cis (\frac{5\pi}{3})$?
|
We normally do exponents before multiplication and the minus sign here is a multiplication. I think the problem wants
$$-(8^{1/3}) \mbox{ non } (-8)^{1/3}.$$
So work out $8^{1/3}$ to get
$$2, 2 \mbox{ cis } \left(\frac{2\pi}{3}\right), 2 \mbox{ cis } \left(\frac{4\pi}{3}\right).$$
Now multiply by $-1$
$$-2, -2 \mbox{ cis } \left(\frac{2\pi}{3}\right), -2 \mbox{ cis } \left(\frac{4\pi}{3}\right)$$
And recall that $-\mbox{ cis } \left(\frac{2\pi}{3}\right)= \mbox{ cis } \left(\frac{5\pi}{3}\right)$ and $-\mbox{ cis } \left(\frac{4\pi}{3}\right) = \mbox{ cis } \left(\frac{\pi}{3}\right).$ so you have
$$-2, 2 \mbox{ cis } \left(\frac{5\pi}{3}\right), 2 \mbox{ cis } \left(\frac{\pi}{3}\right).$$
I believe the right answer has $-2$ rather than $2$.
|
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How $\int_{0}^{\infty} \frac{\arctan(x)}{1+x}\frac{dx}{\sqrt[4]{x}}=\frac{\pi}{\sqrt2}\big(\pi/2+\ln{\beta}\big)$ $$
\int_{0}^{\infty}\frac{\arctan\left(x\right)}{1 + x}
\,\frac{\mathrm{d}x}{\sqrt[{\large 4}]{x}} =
\frac{\pi}{\,\sqrt{\,{2}\,}\,}
\left[{\pi \over 2} + \ln\left(\,{\beta}\,\right)\right]
$$
$$
\mbox{Find the value of}\quad
\beta^{4} - 28\beta^{3} + 70\beta^{2} - 28\beta.
$$
How to do this question ?. I tried conventional approaches such as substituting $x$ with $1/t^{2}$ but none of them is yielding an answer.
|
Write
\begin{align*}
I
:= \int_{0}^{\infty}\frac{\arctan x}{(x+1)x^{1/4}} \, \mathrm{d}x
= \int_{0}^{1}\int_{0}^{\infty}\frac{x^{3/4}}{(x+1)(t^2x^2+1)} \, \mathrm{d}x\mathrm{d}t.
\end{align*}
The inner integral can be computed via a standard technique involving the contour integration along the keyhole contour. Indeed, denoting by $\operatorname{Log}$ the complex logarithm with the branch cut $[0,\infty)$ and writing
$$ f(z) = \frac{\exp\left(\frac{3}{4}\operatorname{Log}(z)\right)}{(z+1)(t^2z^2+1)}, $$
we have
\begin{align*}
\int_{0}^{\infty}\frac{x^{3/4}}{(x+1)(t^2x^2+1)} \, \mathrm{d}x
&= \frac{2\pi i}{1 - e^{3\pi i/2}} \left(
\underset{z=-1}{\mathrm{Res}} \, f(z)
+ \underset{z=i/t}{\mathrm{Res}} \, f(z)
+ \underset{z=-i/t}{\mathrm{Res}} \, f(z) \right) \\
&= -\frac{\pi}{\sin(3\pi/4)} \left( \frac{1}{t^2+1} + \frac{e^{-3i\pi/8}}{2(it-1)t^{3/4}} + \frac{e^{3i\pi/8}}{2(-it-1)t^{3/4}} \right).
\end{align*}
Now by noting that
$$ \int_{0}^{1} \frac{\omega \, \mathrm{d}t}{(\omega^4 t - 1)t^{3/4}}
= \int_{0}^{1} \frac{\omega \, \mathrm{d}u}{(\omega u)^4 - 1}
= -2\arctan(\omega) + \log\left(\frac{1-\omega}{1+\omega}\right) $$
holds for any complex $\omega$ avoiding the branch cuts $\cup_{k=0}^{3} i^k [1,\infty)$, the original integral reduces to
\begin{align*}
I
&= - \pi \sqrt{2} \biggl( \frac{\pi}{4} - \arctan(e^{3i\pi/8}) - \arctan(e^{-3i\pi/8}) \\
&\hspace{5em} + \frac{1}{2}\log\left(\frac{1-e^{3i\pi/8}}{1+e^{3i\pi/8}}\right) + \frac{1}{2}\log\left(\frac{1-e^{-3i\pi/8}}{1+e^{-3i\pi/8}}\right) \biggr) \\
&= - \pi \sqrt{2} \left( \frac{\pi}{4} - \frac{\pi}{2} + \log\tan\left(\frac{3\pi}{16}\right) \right) \\
&= \frac{\pi}{\sqrt{2}}\left( \frac{\pi}{2} - 2\log\tan\left(\frac{3\pi}{16}\right) \right).
\end{align*}
Here, the second line follows from the identities $\arctan z+\arctan(1/z) = \frac{\pi}{2}$ for $\operatorname{Re}(z) > 0$ and $\frac{1-e^{i\theta}}{1+e^{i\theta}}=-i\tan(\frac{\theta}{2})$. This shows that
$$ \beta = \cot^2\left(\frac{3\pi}{16}\right). $$
Finally, using the observation that
$$ X = \beta + \beta^{-1} = \frac{4}{\sin^2(3\pi/8)} -2 = 14 - 8\sqrt{2} $$
is a zero of the equation $X^2 - 28X + 68 = 0$, we get
$$ \beta^4 - 28\beta^3 + 70\beta^2 - 28\beta + 1 = 0, $$
and therefore the answer is $-1$.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3792655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Find $\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$ In what ways can I evaluate $$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
I tried a few methods but none work or simplify things and i cant think of substitutions that could turn things better, while the substitution $x=1/x$ gets
$$\int _0^{\infty }\frac{\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^{\infty }\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
i dont know how to proceed, splitting the integral at point 1 also doesnt help much.
The integral can also be expressed as
$$I=\int _0^1\frac{\left(1+x^2\right)\ln \left(1+x\right)}{1-x^2+x^4}\:\mathrm{d}x-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
that first integral as mentioned in the comments has been evaluated and is
$$I=\frac{\pi }{6}\ln \left(2+\sqrt{3}\right)-\int _0^1\frac{x^2\ln \left(x\right)}{1-x^2+x^4}\:\mathrm{d}x$$
But how to tackle the second one?
|
Split the integration range
\begin{align}
I& =\int _0^{\infty }\frac{\ln (1+x)}{1-x^2+x^4}dx= \int _0^1\frac{(1+x^2)\ln (1+x)-x^2\ln x}{1-x^2+x^4}dx\\
\end{align}
Then, integrate by parts via
$$d\left( \cot^{-1}\frac x{x^2-1}\right)=\frac{1+x^2}{1-x^2+x^4}dx$$
$$d\left( \frac12\tan^{-1}\frac x{1-x^2} - \frac1{2\sqrt3}\tanh^{-1}\frac {\sqrt3x}{1+x^2}\right)= \frac{x^2}{1-x^2+x^4}dx
$$
to express the integral as
\begin{align}
I= I_1 -\frac1{2\sqrt3}I_2 +\frac12I_3\tag1
\end{align}
where
$$
I_1 = \int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {x}{1-x^2},\>\>\>\>\>
I_2 = \int_0^1 \frac{dx}{x} \tanh^{-1}\frac {\sqrt3x}{1+x^2}\\
I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2}
$$
To evaluate $I_1$, let $J_1(a) =\int_0^1 \frac{dx}{1+x} \cot^{-1}\frac {2x\sin a}{1-x^2}$
\begin{align}
J_1’(a) &= \int_0^1 \frac{2\cos a (x-x^2)dx}{(x^2+1)^2-(2x\cos a)^2}= - \frac\pi4\tan\frac a2+\frac12\left( a\>{\csc a}+ \ln\tan\frac a2\right)
\end{align}
\begin{align}
I_1 &=J_1(\frac\pi6) = J_1(0)+\int_0^{\frac\pi6}J_1’(a)da \\
&= \frac\pi2\ln2-\frac\pi4 \int_0^{\frac\pi6}\tan\frac a2 da+\frac12\int_0^{\frac\pi6} d\left( a\ln\tan\frac a2\right)\\
&=\frac\pi2\ln2 -\frac\pi2\ln\cos\frac\pi{12}-\frac\pi{12}\ln\tan\frac\pi{12}= \frac\pi6\ln(2+\sqrt3)\tag2
\end{align}
To evaluate $I_2$, let $J_2(a) =\int_0^1 \frac{dx}{x} \tanh^{-1}\frac {2ax}{1+x^2}$
\begin{align}
J_2’(a) &= \int_0^1 \frac{2 (1+x^2)}{(x^2+1)^2-(2ax)^2}dx = \frac\pi2\frac1{\sqrt{1-a^2}}
\end{align}
\begin{align}
I_2 &=J_2(\frac{\sqrt3}2) = \int_0^{\frac{\sqrt3}2}J_2’(a)da = \frac\pi2 \int_0^{\frac{\sqrt3}2} \frac{da}{\sqrt{1-a^2}}=\frac{\pi^2}6\tag3
\end{align}
To evaluate $I_3$
$$I_3 = \int_0^1 \frac{dx}{x} \tan^{-1}\frac {x}{1-x^2}
= \int_0^1 \frac{\tan^{-1}x}x dx+ \int_0^1\underset{x^3\to x}{\frac{\tan^{-1}x^3}x}dx\\
= \frac43\int_0^1 \frac{\tan^{-1}x}x dx= \frac43G\tag4
$$
Plug (2), (3) and (4) into (1) to obtain
$$I=\frac\pi6\ln(2+\sqrt3) -\frac{\pi^2}{12\sqrt3} +\frac23G
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3793410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Verifying the closed form of $\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}$ My question is at the end of the solution.
We have
$$\int_0^1 x^{n-1}\ln(1-x)=-\frac{H_n}{n}$$
Differentiate both sides with respect to $n$
$$\int_0^1 x^{n-1}\ln x\ln(1-x)=\frac{H_n}{n^2}+\frac{H_n^{(2)}}{n}-\frac{\zeta(2)}{n}$$
Next multiply both sides by $\frac{4n}{n{2n\choose n}}$ then $\sum_{n=1}^\infty$ we get
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\sum_{n=1}^\infty\frac{(4x)^n}{n{2n\choose n}}\right)dx$$
$$=\int_0^1\frac{\ln x\ln(1-x)}{x}\left(\frac{2\sqrt{x}\arcsin\sqrt{x}}{\sqrt{1-x}}\right)dx$$
$$\overset{\sqrt{x}=\sin\theta}{=}16\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta=16I$$
For the integral, let $\theta\to \frac{\pi}{2}-\theta$ to have
$$I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta-\int_0^{\pi/2}\theta\ln(\sin\theta)\ln(\cos\theta)d\theta$$
$$\Longrightarrow 2I=\frac{\pi}{2}\int_0^{\pi/2}\ln(\sin\theta)\ln(\cos\theta)d\theta$$
$$=\frac{\pi}{2}\left(\frac{\pi}{2}\ln^2(2)-\frac{\pi^3}{48}\right)$$
$$\Longrightarrow I=\frac34\ln^2(2)\zeta(2)-\frac{15}{32}\zeta(4)$$
Therefore
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}-\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)-\frac{15}{2}\zeta(4)$$
Since
$$\zeta(2)\sum_{n=1}^\infty\frac{4^n}{n^2{2n\choose n}}=\zeta(2)\left(\frac{\pi^2}{2}\right)=\frac{15}{2}\zeta(4)$$
we have the nice relation
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}+\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=12\ln^2(2)\zeta(2)$$
Finally substitute
$$\sum_{n=1}^\infty\frac{4^nH_n}{n^3{2n\choose n}}=-8\text{Li}_4\left(\frac12\right)+\zeta(4)+8\ln^2(2)\zeta(2)-\frac{1}{3}\ln^4(2)$$
we obtain
$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}=8\text{Li}_4\left(\frac12\right)-\zeta(4)+4\ln^2(2)\zeta(2)+\frac{1}{3}\ln^4(2)\approx 6.2957$$
But Mathematica gives
$$\sum_{n=1}^\infty\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.04326$$
Can you spot any mistake or my solution is good?
Thank you.
|
@User 628759 gave the solution. It is quite surprising to see how sensitive it the result to this parameter.
On my side, I computed exactly
$$\sum_{n=1}^{10000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.23740$$
$$\sum_{n=1}^{20000}\frac{4^nH_n^{(2)}}{n^2{2n\choose n}}\approx 6.25448$$
To be slow, it is !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3793529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$ Question:
The equation $2x^2+ax+(b+3)=0$ has real roots. Find the minimum value of $a^2+b^2$
My working:
$D=a^2-8(b+3)\geqq0$
$a^2\geqq8(b+3)=8b+24$
Add $b^2$ to both sides
$a^2+b^2\geqq b^2+8b+24=(b+4)^2+8\geqq 8$
$\therefore {(a^2+b^2)
}_{min} = 8$
However the answer in the book is 9.
What have I done wrong?
|
By your work: $$a^2+b^2=a^2+b^2-\frac{3}{4}(a^2-8(b+3))+\frac{3}{4}(a^2-8(b+3))\geq$$
$$\geq a^2+b^2-\frac{3}{4}(a^2-8(b+3))=\frac{1}{4}(a^2+4(b+3)^2)+9\geq9.$$
The equality occurs for $a=0$ and $b=-3$, which says that we got a minimal value.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3794277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Apporoaches to solve the given algebraic expression If $\displaystyle \ \ x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$ then what is the value of the given expression
$$\displaystyle \left(\frac{x+1}{x-1}\right)^{\frac{1}{3}} +\ \left(\frac{x-1}{x+1}\right)^{\frac{1}{3}} \ =\ \ ?$$
My Try :
As I can find the value of $\displaystyle x$, from the given equation but it will be tedious I think !.
$$\displaystyle x^{4} \ +\ x^{2} \ =\ \frac{11}{5}$$
$$\displaystyle \Longrightarrow \ x^{2} +1/2 \ =\ \frac{7}{2\sqrt{5}}$$
$$\displaystyle \Longrightarrow \ x^2 \ =\ \ \frac{7-\sqrt{5}}{2\sqrt{5}}$$
Which is getting too much complicated to solve the expression by putting the value of $\displaystyle x$.
What could be the other way to solve the given expression?
|
Consider $A = (\frac{x+1}{x-1})^\frac{1}{3} + (\frac{x-1}{x+1})^\frac{1}{3}$ and $x = \tan u$. Since $x^4+x^2 = \frac{11}{5}$ then $\frac{\sin^2u}{(1-\sin^2u)^2}=\frac{11}{5}$ by solving a simple quadric equation we get that $\sin u = \pm \sqrt{\frac{27-7\sqrt5}{22}}$.
Therefore $A = (\frac{\tan u+1}{\tan u-1})^\frac{1}{3} + (\frac{\tan u+1}{\tan u-1})^\frac{1}{3}$ so from the tangent formula for adding two arc we get: $-A = \tan^\frac{1}{3} {(u+\frac{\pi}{4})}+ \cot^\frac{1}{3} {(u+\frac{\pi}{4})}$. By rasing both sides to 3 we get:$$3A-A^3=\pm\frac{1}{\sin(u+\frac{\pi}{4})\times\sqrt{1-\sin^2{(u+\frac{\pi}{4})}}}$$On the other hand we know: $\sin(u+\frac{\pi}{4})=\pm\frac{\sqrt {225\sqrt 5-280}}{11}$. By simplifications and solving a simple polynomial equation of third degree we can easily get the final answer.
I hope that it can help you. Thank you for your question.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3794954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Show that $f(x) = x|x|$ is continuous and differentiable - solution verification? Another exercise I did without any solutions.
I highly doubt this is correct so pls correct me :)
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be given by $f(x):=x|x| .$ Show that $f$ is continuous and differentiable on $\mathrm{R}$
$$
\begin{array}{l}
\text { Continuous: } \lim _{x \rightarrow c} f(x)=f(c) \\
\begin{aligned}
\lim _{x \rightarrow c} x \cdot|x| &=\lim _{x \rightarrow c} x \cdot \lim _{x \rightarrow c}|x|=f(c) \\
&=\lim _{x \rightarrow c} c \cdot \lim _{x \rightarrow c}|c|=f(c) \\
&=c \cdot|c|=f(c)=c \cdot|c|
\end{aligned}
\end{array}
$$
So $f(x)$ is continuous
Differentiable: show $f^{\prime}(x)$ exists atall $x \in \mathbb{R}$ :
$$
\begin{array}{l}\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\
\lim _{h \rightarrow 0} \frac{(x \cdot|x|)+h-(x \cdot|x|)}{h} \\ =\lim _{h \rightarrow 0} \frac{h}{h}=1\end{array}
$$
$$
So f(x) \text { is differentiable }
$$
|
For $x\neq 0$, $$\begin{align*} \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}&= \lim_{h\to 0} \frac{(x+h)|x+h|-x|x|}{h}\\&=\lim_{h\to 0} \frac{x|x+h| + h|x+h|-x|x|}{h}\\&=\lim_{h\to 0} \frac{x(|x+h|-|x|)+h|x+h|}{h}\\&= \lim_{h\to 0}\frac{x(|x+h|-|x|)}{h} + \frac{h|x+h|}{h}\\&=\bigg[x\lim_{h\to 0}\frac{|x+h|-|x|}{h}\bigg]+|x|\\&=\frac{x^2}{|x|}+|x|\\&= 2\frac{x^2}{|x|}\\&=2\bigg|\frac{x^2}{x}\bigg|\\&=2|x|\end{align*} $$
For the left and right hand limits of $$f'(x)=\frac{x^2}{|x|}+|x|$$ as $x\to 0$, both go to $0$, so $f(x)$ is differentiable at $0$.
Note:For $g(x)=|x|$, $$\begin{align*} g'(x)&=\lim_{h\to 0} \frac{|x+h|-|x|}{h}\\&=\lim_{h\to 0}\frac{\sqrt{(x+h)^2}-\sqrt{x^2}}{h}\\&=\lim_{h\to 0} \frac{(x+h)^2-x^2}{h(\sqrt{(x+h)^2}+\sqrt{x^2})}\\&=\lim_{h\to 0}\frac{x^2+2xh+h^2-x^2}{h(\sqrt{(x+h)^2}+\sqrt{x^2})}\\&= \lim_{h\to 0} \frac{2x+h}{\sqrt{(x+h)^2}+\sqrt{x^2}}\\&= \frac{2x}{2|x|}\\&=\frac{x}{|x|}\end{align*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/3795565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
$ L(P)= (1+X^2)P''(X)-2XP'(X)$
*
*$E=\mathbb{R}[X]$
*$L$ is an endomorphism of $E$ and $L(P)= (1+X^2)P''(X)-2XP'(X)$
What are the possible eigenvalues of $L$ and the dimension of the eigenspace ?
Let's $p(x)=\sum_{k=0}^{n} a_k X^k$ be an eigenvector associated to the eigenvalue $\lambda$.
$a_{n+1} = a_{n+2} = .. =0$
$
\begin{cases}
p(x) &= \sum_{k=0}^{+\infty} a_k x^k \\
p'(x) &= \sum_{k=1}^{+\infty} a_k x^{k-1} k \\
p''(x) &= \sum_{k=2}^{+\infty} a_k x^{k-2} k (k-1) \\
\end{cases}
$
$
\begin{cases}
-2x p'(x) &= \sum_{k=1}^{+\infty} -2 a_k x^k \\
x^2 p''(x) &= \sum_{k=2} a_k k (k-1) x^k \\
p''(x) & = \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) \\
\end{cases}
$
$
\begin{align*}
(1+x^2)P''(x) -2xP'(x) - x P'(x) &= \lambda P(x) \\
\sum_{k=1}^{+\infty} -2 k a_k x^k +\sum_{k=2} a_k k (k-1) x^k + \sum_{k=0}^{+\infty} a_{k+2} x^{k} (k+2) (k+1) &= \lambda \sum_{k=0}^{+\infty} a_k x^k \\
\end{align*}
$
$
\begin{cases}
-2 k a_k + a_k k (k-1) + a_{k+2} (k+2) (k+1) &= \lambda a_k ~~ \text{for} ~ k \geq 2 \\
a_2 \times 2 &= \lambda a_0 \\
-2 a_1 +a_3 \times 6 &= \lambda a_1 \\
\end{cases}
$
\begin{cases}
(k^2-3k-\lambda)a_k &=-(k+2)(k+1)a_{k+2}\tag{1} \\
a_2 \times 2 &= \lambda a_0 \\
-2 a_1 +a_3 \times 6 &= \lambda a_1 \\
\end{cases}
$(1)$ is available only for $k \geq 2$.
$
\begin{align*}
\psi : \mathbb{N} &\to \mathbb{N}\\
n & \mapsto n^2-3n \\
\end{align*}
$
$\psi(n)-\psi(m) = (n-m)(n+m-3)$
|
Let $\lambda$ be an eigenvalue of $L$ and $p(x)=\sum_{k=0}^n a_kx^k=\sum_{k=0}^\infty a_kx^k$ with $a_n=1$ and $0=a_{n+1}=a_{n+2}=\cdots$ be a corresponding eigenvector. From $(1+x^2)p''(x)-2xp'(x)=\lambda p(x)$, we obtain the recurrence relation
$$
\left[(k+2)(k+1)a_{k+2}+k(k-1)a_k\right]-2ka_k=\lambda a_k
$$
or equivalently
$$
(k^2-3k-\lambda)a_k=-(k+2)(k+1)a_{k+2}\tag{1}
$$
for each $k\ge0$. It follows that $\lambda=n^2-3n$.
Since $(n^2-3n)-(m^2-3m)=(n-m)(n+m-3)$, if $\lambda=n^2-3n$ for some $n>3$, then $n$ is uniquely determined and the LHS of $(1)$ is nonzero for every $0\le k<n$. Thus $a_{n-1},a_{n-2},\ldots,a_0$ can be uniquely and recursively determined from $(1)$ using the boundary conditions that $a_{n+1}=0$ and $a_n=1$. Hence the eigenspace for $\lambda$ exists and is one-dimensional.
When $\lambda=n(n-3)$ for some $0\le n\le3$, complications arise because $\lambda$ has two factorisations $n(n-3)$ and $m(m-3)$, where $m=3-n$.
*
*When $\lambda=0$, the equation $\lambda=n(n-3)=0$ has two nonnegative integer solutions $n=3$ and $n=0$. The null space of $L$ therefore consists of only cubic polynomials and constant polynomials. Suppose $p$ be a monic cubic polynomial in the null space of $L$. The recurrence relation $(1)$ then gives $a_2=0$ and $a_1=3$, but $a_0$ is undetermined because $(1)$ reduces to $0a_0=0$ when $k=0$. Hence the null space of $L$ exists and it is a two-dimensional subspace spanned by $x^3+3x$ and $1$.
*When $\lambda=-2$, the equation $\lambda=n(n-3)$ has two nonnegative integer solutions $n=2$ and $n=1$. The corresponding eigenspace thus consists of (apart from the zero polynomial) only polynomials of degrees $1$ and $2$. Let $p$ be a monic quadratic polynomial in the eigenspace. The recurrence relation $(1)$ gives $a_0=1$ but $a_1$ is undetermined because $(1)$ reduces to $0a_1=0$ when $k=1$. Therefore the eigenspace for this eigenvalue exists and it is a two-dimensional subspace spanned by $x^2-1$ and $x$.
In short, the eigenvalues of $L$ are $\lambda=n^2-3n$ for each $n\ge2$. The corresponding eigenspace is two-dimensional when $n=2,3$ and one-dimensional when $ n>3$.
|
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"url": "https://math.stackexchange.com/questions/3795961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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|
Does $(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) $ hold? Does there exist a positive constant $c>0$ such that
$$(x-1)^2+(y-1)^2 \le c\big((x-y)^2+(xy-1)^2\big) \tag{1}$$
holds for any nonnegative $x,y$?
Let me add some context for this question:
The motivation comes from the case were $x,y$ are interpreted as singular values of a $2 \times 2$ matrix $A$ with nonnegative determinant. Then $f(x,y)=(x-1)^2+(y-1)^2=\operatorname{dist}^2(A,\operatorname{SO}(2))$.
I am interested in bounding $\operatorname{dist}^2(A,\operatorname{SO}(2))$ from above by a sum of two terms: a term which penalizes deviations of $A$ from being area-preserving, and a term $\operatorname{dist}^2(A,\operatorname{CO}(2))$, which penalizes deviations from being conformal. (Here $\operatorname{CO}(2)=R^{+}\operatorname{SO}(2)$ is the group of conformal matrices).
In an answer to this previous question of mine, the following bound was proved:
$$
(x-1)^2+(y-1)^2 \le |x-y||x+y| + 2|xy-1|.
$$
While this is close to what I had in mind, the term $|x-y||x+y|$ can be large even when $x,y$ become very close. In fact, one can prove that $\operatorname{dist}^2(A,\operatorname{CO}(2))=\frac{1}{2}(\sigma_1(A)-\sigma_2(A))^2$, so this is the reason for asking about the specific bound $(1)$. (The term $(x-y)^2$ corresponds to $\operatorname{dist}^2(A,\operatorname{CO}(2))$).
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Let $x=y=0,$ we get $c \geqslant 2.$
For $c =2,$ inequality become
$$(x-1)^2+(y-1)^2 \leqslant 2\big((x-y)^2+(xy-1)^2\big),$$
equivalent to
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8xy.$$
Using the AM-GM inequality, we have
$$2x^2y^2+x^2+y^2+2(x+y) \geqslant 8\sqrt[8]{(x^2y^2)^2 \cdot x^2 \cdot y^2 \cdot (xy)^2}=8xy.$$
So, your inequality holds for all $c \geqslant 2.$
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Solving $\sqrt{9-x^2} > x^2 + 1$ without graphic calculator for the exact form
Is there any way to solve this inequality without using a graphic calculator to get the exact form?
$$\sqrt{9-x^2} > x^2 + 1$$
I've tried completing the square but I end up with
$$\frac{3 - \sqrt{41}}{2} < x^2 < \frac{3 + \sqrt{41}}{2}$$
which does not match with the answer on Desmos.
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$$9-x^2>x^4+2x^2+1$$
$$x^4+3x^2-8<0$$
$$\left(x^2+\frac32 \right)^2 < 8+\frac94$$
$$\left(x^2+\frac32 \right)^2< \frac{41}4$$
$$0\le x^2<\color{red}-\frac32 + \frac{\sqrt{41}}2$$
Now, your answer should coincide.
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|
Solve $\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$ without L'Hôpital's rule How would you solve the limit
$$\lim_{x\rightarrow +\infty}\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right) = 0$$
without using L'Hôpital's rule?
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$$\frac{1}{x}\log\left(\frac{x+1}{1+x^2}\right)=\frac{1}{x} \log \frac{1}{x}+\frac{\log \left(1+\frac{1}{x} \right)}{\frac{1}{x}\cdot x^2}-\frac{\log \left(1+\frac{1}{x^2} \right)}{\frac{1}{x^2}\cdot x^3}$$
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Proving $\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$ For $a,b,c\ge 0$ Prove that $$\frac{a^3+b^3+c^3}{3}-abc\ge \frac{3}{4}\sqrt{(a-b)^2(b-c)^2(c-a)^2}$$
My Attempt WLOG $b=\text{mid} \{a,b,c\},$
$$\left(\dfrac{a^3+b^3+c^3}{3}-abc\right)^2-\dfrac{9}{16}(a-b)^2(b-c)^2(c-a)^2$$
\begin{align*} &=\frac{1}{9}(a+b+c)^2(a-2b+c)^4\\ &+\frac{2}{3}(a+b+c)^2(a-2b+c)^2(b-c)(a-b)\\ &+\frac{1}{16}(a-b)^2(b-c)^2(a+4b+7c)(7a+4b+c)\\&\geqslant 0\end{align*}
However, this solution is too hard for me to find without computer. Could you help me with figuring out a better soltuion?
Thank you very much
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Let $a=\min\{a,b,c\}$, $b=a+u,$ $c=a+v$ and $u^2+v^2=2tuv$.
Thus, by AM-GM $t\geq1$ and we need to prove that:
$$2(a+b+c)\sum_{cyc}(a-b)^2\geq9\sqrt{\prod_{cyc}(a-b)^2}$$ or
$$2(3a+u+v)(u^2+v^2+(u-v)^2)\geq9\sqrt{u^2v^2(u-v)^2},$$ for which it's enough to prove that
$$4(u+v)(u^2-uv+v^2)\geq9\sqrt{u^2v^2(u-v)^2}$$ or
$$16(u+v)^2(u^2-uv+v^2)^2\geq81u^2v^2(u-v)^2$$ or
$$16(t+1)(2t-1)^2\geq81(t-1)$$ or
$$64t^3-129t+97\geq0,$$ which is true by AM-GM:
$$64t^3+97=64t^3+2\cdot\frac{97}{2}\geq3\sqrt[3]{64t^3\cdot\left(\frac{97}{2}\right)^2}>129t.$$
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Prove that $\tan(x)\tan(x+\frac{\pi}{3})+\tan(x)\tan(\frac{\pi}{3}-x)+\tan(x+\frac{\pi}{3})\tan(x-\frac{\pi}{3}) = -3$ Let's assume that $\tan(x) = y$.
So, $\tan\Big(x+\dfrac{\pi}{3}\Big) = \dfrac{\tan(x) + \tan\Big(\dfrac{\pi}{3}\Big)}{1-\tan(x)\tan\Big(\dfrac{\pi}{3}\Big)} = \dfrac{y+\sqrt{3}}{1-\sqrt{3}y}$
Similarly, $\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{y-\sqrt{3}}{1+\sqrt{3}y}$
Also, $\tan\Big(\dfrac{\pi}{3}-x\Big) = -\tan\Big(x-\dfrac{\pi}{3}\Big) = \dfrac{\sqrt{3}-y}{1+\sqrt{3}y}$
Now, $\tan(x)\tan\Big(x+\dfrac{\pi}{3}\Big)+\tan(x)\tan\Big(\dfrac{\pi}{3}-x\Big)+\tan\Big(x+\dfrac{\pi}{3}\Big)\tan\Big(x-\dfrac{\pi}{3}\Big)$
$$ = y\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)+y\Big(\dfrac{\sqrt{3}-y}{1+\sqrt{3}y}\Big)+\Big(\dfrac{y+\sqrt{3}}{1-\sqrt{3}y}\Big)\Big(\dfrac{y-\sqrt{3}}{1+\sqrt{3}y}\Big)$$
$$ = y\Big(\dfrac{(y+\sqrt{3})(1+\sqrt{3}y)+(1-\sqrt{3}y)(\sqrt{3}-y)}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$
$$ = y\Big(\dfrac{y+\sqrt{3}y^2+\sqrt{3}+3y+\sqrt{3}-y-3y+\sqrt{3}y^2}{1-3y^2}\Big)+\Big(\dfrac{y^2-3}{1-3y^2}\Big)$$
$$ = \dfrac{2\sqrt{3}y+2\sqrt{3}y^3+y^2-3}{1-3y^2}$$
This is how much I've been able to simplify the expression but I'm unable to continue. I am familiar with the values of trigonometric functions at multiples and sub multiples of angles and I think the solution would involve their use (as the question has been taken from that very chapter).
Thanks!
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Like Proving a fact: $\tan(6^{\circ}) \tan(42^{\circ})= \tan(12^{\circ})\tan(24^{\circ})$
$$t_0=\tan y,t_1=\tan\left(y-\dfrac\pi3\right),t_2=\tan\left(y+\dfrac\pi3\right)$$ are the roots of
$$t^3-(3\tan3y)t^2-3t+3\tan3y=0$$
Using Vieta's formula $$t_0t_1+t_1t_2+t_2t_0=\dfrac{-3}1$$
$$t_0+t_1+t_2=\dfrac{3\tan3y}1$$
$$t_0t_1t_2=\dfrac{-3\tan3y}1$$
if $t_0,t_1,t_2$ are finite i.e., $y+\dfrac{n\pi}3\ne m\pi+\dfrac\pi2; n=-1,0,1$
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Solving the recurrence relation $T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$ I have been trying to solve the following recurrence relation
$T(n) = \frac{1}{n}(T(0) + T(1) + ... + T(n-1)) + 5n$
$T(0) = 0$
I've tried to use substitution which wasn't very useful as I couldn't figure out a way to simplify the resultant equation.
My next approach was to just plug in numbers as follows
$T(1) = 5 * 1$
$T(2) = \frac{1}{2}(5 * 1) + 5 * 2 = \frac{5 * 1}{2} + 5 * 2$
$T(3) = \frac{1}{3}(\frac{5 * 1}{2} + 5 * 2) + 5 * 3 = \frac{5 * 1}{6} + \frac{5 * 2}{3} + 5 * 3$
$T(4) = \frac{1}{4}(\frac{5 * 1}{6} + \frac{5 * 2}{3} + 5 * 3) + 5 * 4 = \frac{5 * 1}{24} + \frac{5 * 2}{12} + \frac{5 * 3}{4} + 5 * 4$
I can sort of see a pattern emerge here like $\sum_{i=1}^{n} \frac{5*i}{previousDenominator/ i}$ however I couldn't really see a way to get that bottom part into an actual equation.
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The calculation should be continued: We get
$ \begin{array}{rcl}
T_n&=&T_0+10n- 5\sum\limits_{k=0}^{n-1}\frac{1}{k+1} \\
&=&T_0+10n- 5 H_n
\end{array} $
Here the $H_n=1+\frac{1}{2}+\cdots +\frac{1}{n}$ are the so called harmonic numbers. It is known that
$H_n= \ln n+\gamma + \varepsilon_n$ where $(\varepsilon_n)_{n\geq 1}$ is a sequence monotonically
decreasing to 0 and $\gamma\approx 0.57721. . .$ is the Euler Mascheroni constant.
Actually therefore one knows there is not what would be called a `closed form solution' for the recurrence; but very good approximations:
$T_n=T_0+10n -5\ln n-5\gamma -5\varepsilon_n.$
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|
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
Given that $a,b,c$ satisfy the equation $x^3-2007 x +2002=0$, then find $\frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1}$
The concept of transformation of roots can be applied here.
So replace
$$x \to \frac{x-1}{x+1}$$
After considerable algebra, a cubic would be obtained, and then the sum can easily be found out
My question: This method is touted around a lot, but I never around how it actually works, especially in this case. If $\frac{x-1}{x+1} =\alpha$, then the transformed equation would be $\alpha^3 -2007 \alpha +2002$, and the sum of all $\alpha$ should still be $2007$. Can I get an explanation for this method?
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Since the two posted Answers so far (that provide a concrete value of the indicated sum) disagree, I'll pursue the approach I'd suggested in a Comment.
First form an equation whose three roots are $y = (x+1)$ instead of $x$. That is:
$$ (y-1)^3 - 2007(y-1) + 2002 = 0 $$
$$ y^3 - 3y^2 - 2004y + 4008 = 0 $$
An equation whose three roots are instead $z = 1/y = 1/(x+1)$ is then:
$$ 4008z^3 - 2004z^2 - 3z + 1 = 0 $$
This is obtained by reversing the order of coefficients. By inspection we have the sum of those three roots:
$$ z_1+z_2+z_3 = \frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} = \frac{2004}{4008} = \frac{1}{2} $$
Finally:
$$ \frac{a-1}{a+1}+\frac{b-1}{b+1} +\frac{c-1}{c+1} = 3 - 2\left(\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1}\right) = 2 $$
It may seem counterintuitive that, in order to transform the roots of a polynomial equation $p(x) = 0$, we make a substitution using the functional inverse of that transform. Let's sketch why this is so, and then do a simple example.
If we wanted instead of the original roots of $p(x)$, some function $f$ applied to each of those roots, consider what equation these transformed roots must satisfy. That is, where $x$ satisfies $p(x) = 0$, what does the transformed value $\alpha = f(x)$ satisfy? Clearly if the functional inverse $f^{-1}$ exists, we can say:
$$ p(f^{-1}(f(x))) = 0 $$
In other words the transformed value $\alpha$ satisfies:
$$ p(f^{-1}(\alpha) = 0 $$
Depending on how complicated $f^{-1}$ is, the substitution may cause us to do some careful algebra to simplify the new equation. So to illustrate, let's do an easy example.
Define the polynomial $p(x) := x^2 - 2x$, which has roots for $p(x) = 0$ at $x = 0,2$.
If we wanted to transform those roots by adding one, then we need to substitute into $p(x)$ the opposite transform, subtracting one. Then:
$$ p(x-1) = (x-1)^2 - 2(x-1) = x^2 - 4x + 3 = (x-1) (x-3) $$
Now we have roots of $p(x-1)$ at $x = 1,3$. In other words we have transformed the roots of $p(x)$ by adding one to each of them, as we wished.
Note that this was essentially the first substitution I used in my solution, to get an equation with roots $a+1,b+1,c+1$.
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|
If $a = \frac{1+\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$? My question is this:
If $a = \frac{1+\sqrt5}2,\frac{1-\sqrt5}2$, then what is $a^{18} + \frac{323}{a^6}$?
It is an AMC style question and is timed, so I will not be able to use solutions with a lot of case work.
I defined $a$ from part a of a 2 part question and would like to know how to do this without the use of a calculator.
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Okay, actually we can do better than what I suggested in the comments. The basic move is to repeatedly use the identity $a^2 = a + 1$ together with binary exponentiation. We have
$$a^3 = a(a^2) = a(a + 1) = a^2 + a = 2a + 1$$
$$a^6 = (a^3)^2 = (2a + 1)^2 = 4a^2 + 4a + 1 = 8a + 5$$
$$a^{12} = (a^6)^2 = (8a + 5)^2 = 64a^2 + 80a + 25 = 144a + 89$$
$$a^{18} = a^6 \cdot a^{12} = (8a + 5)(144a + 89) = 1152a^2 + 1432a + 445 = 2584a + 1597$$
which tells us the first term. To evaluate the second term it's convenient to work as follows. Introduce $b = \frac{1 - \sqrt{5}}{2}$, the conjugate of $a$, which satisfies $ab = -1$ and $a + b = 1$. We have $\frac{323}{a^6} = 323b^6$, and $b$ also satisfies $b^2 = b + 1$, so exactly the same computation as before applies and we get $b^6 = 8b + 5 = 13 - 8a$. Altogether this gives
$$a^{18} + \frac{323}{a^6} = (2584a + 1597) + 323(13 - 8a) = \boxed{5796}$$
if I haven't made any arithmetic errors.
Some discussion. In general we can prove by induction that
$$a^n = F_n a + F_{n-1}$$
where $F_n$ are the Fibonacci numbers. Binary exponentiation applied to $a$ then proves a doubling identity for the Fibonacci numbers allowing them to be calculated quickly in the same way that binary exponentiation allows powers to be calculated quickly. Specifically we get
$$a^{2n} = F_{2n} a + F_{2n-1} = (F_n a + F_{n-1})^2 = F_n^2 a^2 + 2 F_n F_{n-1} a + F_{n-1}^2 = (F_n^2 + 2 F_n F_{n-1}) a + (F_n^2 + F_{n-1}^2)$$
which gives the pair of doubling identities
$$F_{2n} = F_n(F_n + 2F_{n-1}) = F_n(F_n + F_{n+1})$$
$$F_{2n-1} = F_n^2 + F_{n-1}^2.$$
This isn't a special feature of $a$ and similar identities can be proven for any quadratic irrational. Abstractly we're repeatedly using the fact that the corresponding quadratic field has basis $\{ 1, a \}$ as a vector space.
The exact same proof conjugated gives that $b^n = F_n b + F_{n-1}$ and subtracting these two identities from each other gives $a^n - b^n = (a - b) F_n$, or slightly rearranging, Binet's formula
$$F_n = \frac{a^n - b^n}{a - b}.$$
There's a more general story to tell here about any sequence defined by a linear recurrence relation; I don't know a good self-contained reference off the top of my head unfortunately.
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Is my proof that $\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n)$ correct? Prove that: $$\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n) $$
$$\frac{n^3-3n^2+2n}{n^2-4n}≥\frac{n^3-3n^2+2n-17}{n^2-4n+22} $$
now we choose$$ N≥6$$
thus
$$\frac{n^3-3n^2+3n^2-18n+2n}{n^2-4n}≥\frac{n^3-3n^2+2n}{n^2-4n}$$
and
$$\frac{n^3-16n}{n^2-4n}=\frac{n^2-16}{n-4}=\frac{(n-4)(n+4)}{n-4}=n+4$$
and if we pick $$c=2$$
we see that $$2n≥n+4$$
for all $$ N≥6$$
and thus
$$\frac{n^3-3n^2+2n-17}{n^2-4n+22}=O(n) $$
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First a small notation error, you define $N$ but don't use it. I think by $N\ge6$, you mean $n\ge N$ for $N=6$.
To prove $a_n = O(n)$ you need to prove $|a_n| \le Cn$ for $n>N$, but you have proven $a_n\le Cn$. You should also have a separate check that for $n\ge6$ (or any larger number) that $a_n\ge 0$. Then $a_n=|a_n|$. Also worth mentioning $n^2-4n+22$ has no integer roots.
Alternate proof:
$$ \left|\frac{n^3-3n^2+2n-17}{n^2-4n+22}\right|\le \frac{n^3+3n^3+2n^3+17n^3}{n^2|1-\frac4n+\frac{22}n|} = \frac{23n}{|1-\frac4n+\frac{22}n|}$$
for $n>100$ (for example), $\frac 4n<\frac14$ and $\frac{22}n<1/4$ so $1-\frac4n +\frac{22}n > \frac12$. Thus
$$\frac{n^3-3n^2+2n-17}{n^2-4n+22} \le 46n$$
so $\frac{n^3-3n^2+2n-17}{n^2-4n+22}= O(n)$.
My constant is much worse and I proved it holds for $n$ much larger only, but these are things allowed by the $O(n)$ notation.
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